Question 13 Marks
Evaluate the following limits: $\lim _{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]$
Answer$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\log x}$
put $x-1=\mathrm{h}$
$\therefore \quad x=1+\mathrm{h}$
As $x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\therefore$ Required limit
$=\lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{\log (1+h)}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\sqrt{1+\mathrm{h}}-1}{\log (1+\mathrm{h})} \times \frac{\sqrt{1+\mathrm{h}}+1}{\sqrt{1+\mathrm{h}}+1} \ldots [$ By rationalization $]$
$=\lim _{h \rightarrow 0} \frac{(1+h)-1}{\log (1+h)} \times \frac{1}{\sqrt{1+h}+1}$
$=\lim _{h \rightarrow 0} \frac{h}{\log (1+h)} \times \frac{1}{\sqrt{1+h}+1}$
$=\lim _{h \rightarrow 0} \frac{1}{\frac{\log (1+h)}{h}} \times \frac{1}{\sqrt{1+h}+1}$
$\ldots\left[\begin{array}{l} \text { Divide Numerator and } \\ \text { Denominator by } \\
\text { Ash } \rightarrow 0, h \neq 0
\end{array} \right]$
$=\lim _{h \rightarrow 0} \frac{1}{\frac{\log (1+h)}{h}} \times \frac{1}{\lim _{h \rightarrow 0}(\sqrt{1+h}+1)}$
$=\frac{1}{1} \times \frac{1}{\sqrt{1+0}+1}$
$=\frac{1}{2}$
View full question & answer→Question 23 Marks
Evaluate the following limits: $\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}\right]$
Answer$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}$
$=\lim _{x \rightarrow 1} \frac{2^{2(x-1)}-2^x+1}{\sin ^2(x-1)}$
$\text { Put } x=1+\mathrm{h}, \therefore x-1=\mathrm{h}$
$\text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=\lim _{h \rightarrow 0} \frac{2^{2 h}-2^{1+h}+1}{\sin ^2 h}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{2^{2 \mathrm{~h}}-2.2^{\mathrm{h}}+1}{\sin ^2 \mathrm{~h}}$
$=\lim _{h \rightarrow 0} \frac{\left(2^h-1\right)^2}{\sin ^2 h}$
$=\lim _{h \rightarrow 0}\left[\frac{\frac{\left(2^h-1\right)^2}{h^2}}{\frac{\sin ^2 h}{h^2}}\right] \ldots[\because h \rightarrow 0, h \neq 0]$
$=\frac{\left(\lim _{h \rightarrow 0} \frac{2^h-1}{h}\right)^2}{\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right)^2}$
$=\frac{(\log 2)^2}{(1)^2}$
$\cdots\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
$=(\log 2)^2$
View full question & answer→Question 33 Marks
Evaluate the following limits:
$\lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^2(7 x-3)^3}{(5 x+2)^5}\right]$
Answer$ \lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^2 \cdot(7 x-3)^3}{(5 x+2)^5}\right]$
$=\lim _{x \rightarrow \infty} \frac{\frac{(2 x+1)^2}{x^2} \cdot \frac{(7 x-3)^3}{x^3}}{\frac{(5 x+2)^5}{x^5}} $
...[Divide Numerator and Denominator by $x^5$ ]
$ =\frac{\lim _{x \rightarrow \infty}\left(\frac{2 x+1}{x}\right)^2 \cdot\left(\frac{7 x-3}{x}\right)^3}{\lim _{x \rightarrow \infty}\left(\frac{5 x+2}{x}\right)^5}$
$=\frac{\lim _{x \rightarrow \infty}\left(2+\frac{1}{x}\right)^2 \cdot \lim _{x \rightarrow \infty}\left(7-\frac{3}{x}\right)^3}{\lim _{x \rightarrow \infty}\left(5+\frac{2}{x}\right)^5}$
$=\frac{(2+0)^2 \cdot(7-0)^3}{(5+0)^5} \quad \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, \mathrm{k}>0\right]$
$=\frac{(2)^2 \times(7)^3}{(5)^5}=\frac{1372}{3125} $
View full question & answer→Question 43 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{\left(5^x-1\right)^2}{\left(2^x-1\right) \log (1+x)}\right]$
Answer$ \lim _{x \rightarrow 0} \frac{\left(5^x-1\right)^2}{\left(2^x-1\right) \cdot \log (1+x)}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(5^x-1\right)^2}{x^2}}{\frac{\left(2^x-1\right) \cdot \log (1+x)}{\dot{x}^2 }} $
... [Divide Numerator and Denominator by $x^2 \because$
$\left.x \rightarrow 0, x \neq 0 \therefore x^2 \neq 0\right]$
$ =\frac{\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)^2}{\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right) \cdot \frac{\log (1+x)}{x}}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{5^x-1}{x}\right)^2}{\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right) \cdot \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}}$
$=\frac{(\log 5)^2}{\log 2 \times 1}$
$\ldots\left[\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log \text { a, } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
$=\frac{(\log 5)^2}{\log 2} $
View full question & answer→Question 53 Marks
Evaluate the following limits:$
\lim _{x \rightarrow 1}\left[\frac{a b^x-a^x b}{x^2-1}\right]
$
Answer$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{a b^x-\mathrm{a}^x \mathrm{~b}}{x^2-1} & =\lim _{x \rightarrow 1} \frac{\mathrm{ab}\left(\mathrm{b}^{x-1}-\mathrm{a}^{x-1}\right)}{x^2-1^2} \\
& =\lim _{x \rightarrow 1} \frac{\mathrm{ab}\left(\mathrm{b}^{x-1}-\mathrm{a}^{x-1}\right)}{(x-1)(x+1)}
\end{aligned}
$
Put $x=1+\mathrm{h}$,
$\therefore \quad x-1=\mathrm{h}$
As $x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=\lim _{h \rightarrow 0} \frac{a b\left(b^h-a^h\right)}{h(1+h+1)}$
$=a b \lim _{h \rightarrow 0} \frac{b^h-1+1-a^h}{h(2+h)}$
$=a b \lim _{h \rightarrow 0} \frac{\left(b^h-1\right)-\left(a^h-1\right)}{h(2+h)}$
$=a b \lim _{h \rightarrow 0} \frac{1}{2+h}\left(\frac{b^h-1}{h}-\frac{a^h-1}{h}\right)$
$=a b \cdot \frac{1}{\lim _{h \rightarrow 0}(2+h)}\left(\lim _{h \rightarrow 0} \frac{b^h-1}{h}-\lim _{h \rightarrow 0} \frac{a^h-1}{h}\right)$
$=$ ab. $\frac{1}{2+0} \cdot(\log \mathrm{b}-\log \mathrm{a}) \cdots\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
$=\frac{a b}{2} \log \left(\frac{b}{a}\right)$
View full question & answer→Question 63 Marks
Evaluate the following limits:$
\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{x-a}\right]
$
Answer$
\lim _{x \rightarrow a} \frac{\sin x-\sin a}{x-a}
$
Put $x=\mathrm{a}+\mathrm{h}$,
$
\therefore \quad x-\mathrm{a}=\mathrm{h}
$
As $x \rightarrow \mathrm{a}, \mathrm{h} \rightarrow 0$
$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow \mathbf{a}} \frac{\sin x-\sin a}{x-a} \\
& =\lim _{h \rightarrow 0} \frac{\sin (a+h)-\sin a}{h} \\
& =\lim _{\mathbf{h} \rightarrow 0} \frac{2 \cos \left(\frac{a+h+a}{2}\right) \sin \left(\frac{a+h-a}{2}\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{2 \cos \left(a+\frac{h}{2}\right) \sin \frac{h}{2}}{h}\\
& =\lim _{h \rightarrow 0} \cos \left(a+\frac{h}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)} \\
& =\cos (a+0)(1) \\
& =\cos a
\end{aligned}
$
View full question & answer→Question 73 Marks
Evaluate the following limits:$
\lim _{x \rightarrow 0}\left[\frac{e^x+e^{-x}-2}{x \cdot \tan x}\right]
$
Answer$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\mathrm{e}^x+\mathrm{e}^{-x}-2}{x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\mathrm{e}^x+\frac{1}{\mathrm{e}^x}-2}{x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\left(\mathrm{e}^x\right)^2+1-2\left(\mathrm{e}^x\right)}{\mathrm{e}^x \cdot x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\left(\mathrm{e}^x-1\right)^2}{\mathrm{e}^x \cdot x \tan x} \\
= & \lim _{x \rightarrow 0} \frac{\left[\frac{\left(\mathrm{e}^x-1\right)^2}{x^2}\right]}{\left(\frac{\mathrm{e}^x \cdot x \tan x}{x^2}\right)} \quad \ldots[\because x \rightarrow 0 ; x \neq 0] \\
= & \lim _{x \rightarrow 0} \frac{\left(\frac{\mathrm{e}^x-1}{x}\right)^2}{\mathrm{e}^x \cdot\left(\frac{\tan x}{x}\right)} \\
= & \frac{(1)^2}{\mathrm{e}^0 \cdot 1} \\
= & 1 \\
= & \left.\lim _{x \rightarrow 0} \mathrm{e}^x \cdot \lim _{x \rightarrow 0} \frac{\mathrm{e}^x-1}{x}\right)^2
\end{aligned}
$
View full question & answer→Question 83 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{\sec x^2-1}{x^4}\right]$
Answer$\lim _{x \rightarrow 0} \frac{\sec x^2-1}{x^4}$
Put $x^2=y$
As $x \rightarrow 0, x^2 \rightarrow 0$
$\therefore \quad y \rightarrow 0$
$\therefore \quad$ Required limit
$ \quad=\lim _{y \rightarrow 0} \frac{\sec y-1}{y^2}$
$\quad=\lim _{y \rightarrow 0} \frac{\sec y-1}{y^2} \times \frac{\sec y+1}{\sec y+1}$
$\quad=\lim _{y \rightarrow 0} \frac{\sec ^2 y-1}{y^2(\sec y+1)}$
$=\lim _{y \rightarrow 0} \frac{\tan ^2 y}{y^2(\sec y+1)}$
$=\lim _{y \rightarrow 0} \frac{\tan ^2 y}{y^2} \times \frac{1}{\sec y+1}$
$=\lim _{y \rightarrow 0}\left(\frac{\tan ^2 y}{y^2}\right) \times \lim _{y \rightarrow 0} \frac{1}{\sec y+1}$
$=(1)^2 \times \frac{1}{1+1}$
$=\frac{1}{2} $
View full question & answer→Question 93 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{x}{|x|+x^2}\right]$
Answer$ \lim _{x \rightarrow 0}\left[\frac{x}{|x|+x^2}\right]$
$|x|=x ; x \geq 0$
$=-x ; x<0$
$\lim _{x \rightarrow 0^{-}}\left[\frac{x}{|x|+x^2}\right]=\lim _{x \rightarrow 0} \frac{x}{-x+x^2}$
$=\lim _{x \rightarrow 0} \frac{x}{x(-1+x)}$
$=\lim _{x \rightarrow 0} \frac{x}{-1+x} \ldots(\text { As } x \rightarrow 0, x \neq 0)$
$=\frac{1}{-1+0}$
$=-1$
$\lim _{x \rightarrow 0^{+}}\left[\frac{x}{|x|+x^2}\right]=\lim _{x \rightarrow 0} \frac{x}{x+x^2}$
$=\lim _{x \rightarrow 0} \frac{x}{x(1+x)}$
$=\lim _{x \rightarrow 0} \frac{1}{1+x} \ldots(\text { As } x \rightarrow 0, x \neq 0)$
$=\frac{1}{1+0}$
$=1$
$\therefore \quad \lim _{x \rightarrow 0^{-}}\left[\frac{x}{|x|+x^2}\right] \neq \lim _{x \rightarrow 0}\left[\frac{x}{|x|+x^2}\right]$
$\therefore \quad \lim _{x \rightarrow 0} \frac{x}{|x|+x^2} \text { does not exist. }$
View full question & answer→Question 103 Marks
Evaluate the following limits: $\lim _{x \rightarrow \infty}\left[\frac{(2 x-1)^{20}(3 x-1)^{30}}{(2 x+1)^{50}}\right]$
Answer$\lim _{x \rightarrow \infty} \frac{(2 x-1)^{20}(3 x-1)^{30}}{(2 x+1)^{50}}$
$=\lim _{x \rightarrow \infty} \frac{\frac{(2 x-1)^{20}}{x^{20}} \cdot \frac{(3 x-1)^{30}}{x^{30}}}{\frac{(2 x+1)^{50}}{x^{30}}}$
$\ldots\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x^{30} \end{array}\right]$
$=\frac{\lim _{x \rightarrow \infty}\left(\frac{2 x-1}{x}\right)^{20} \cdot\left(\frac{3 x-1}{x}\right)^{30}}{\lim _{x \rightarrow \infty}\left(\frac{2 x+1}{x}\right)^{50}}$
$=\frac{\lim _{x \rightarrow \infty}\left(2-\frac{1}{x}\right)^{20} \cdot \lim _{x \rightarrow \infty}\left(3-\frac{1}{x}\right)^{30}}{\lim _{x \rightarrow \infty}\left(2+\frac{1}{x}\right)^{50}}$
$=\frac{(2-0)^{20} \times(3-0)^{30}}{(2+0)^{50}} \cdots\left[\lim _{x \rightarrow \infty} \frac{1}{x^{ k }}=0, k >0\right]$
$=\frac{(2)^{20} \times(3)^{30}}{(2)^{50}}$
$=\frac{(3)^{30}}{(2)^{30}}$
$=\left(\frac{3}{2}\right)^{30}$
View full question & answer→Question 113 Marks
Evaluate the following limits: $\lim _{x \rightarrow \infty}\left[\frac{(3 x-4)^3(4 x+3)^4}{(3 x+2)^7}\right]$
Answer$\lim _{x \rightarrow \infty}\left[\frac{(3 x-4)^3(4 x+3)^4}{(3 x+2)^7}\right] \\
=\lim _{x \rightarrow \infty} \frac{(3 x-4)^3 \frac{(4 x+3)^4}{x^3}}{\frac{(3 x+2)^7}{x^7}} \cdots\left[\begin{array}{l}
\text { Divide numerator and } \\
\text { denominator by } x^7
\end{array}\right]$
$=\frac{\lim _{x \rightarrow \infty}\left(\frac{3 x-4}{x}\right)^3\left(\frac{4 x+3}{x}\right)^4}{\lim _{x \rightarrow \infty}\left(\frac{3 x+2}{x}\right)^7}$
$=\frac{\lim _{x \rightarrow \infty}\left(3-\frac{4}{x}\right)^3 \times \lim _{x \rightarrow \infty}\left(4+\frac{3}{x}\right)^4}{\lim _{x \rightarrow \infty}\left(3+\frac{2}{x}\right)^7}$
$=\frac{(3-0)^3 \times(4+0)^4}{(3+0)^7} \ldots\left[
\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k >0
\right]$
$=\frac{(3)^3 \times(4)^4}{(3)^7}$
$=\frac{(4)^4}{81}$
View full question & answer→Question 123 Marks
Evaluate the following limits: $\lim _{x \rightarrow \infty}\left[\sqrt{x^4+4 x^2}-x^2\right]$
Answer$\lim _{x \rightarrow \infty}\left[\sqrt{x^4+4 x^2}-x^2\right]$
$=\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^4+4 x^2}-x^2\right)\left(\sqrt{x^4+4 x^2}+x^2\right)}{\sqrt{x^4+4 x^2}+x^2}$
$[$By rationalizatio$]$
$=\lim _{x \rightarrow \infty} \frac{x^4+4 x^2-x^4}{\sqrt{x^4+4 x^2}+x^2}$
$=\lim _{x \rightarrow \infty} \frac{4 x^2}{\sqrt{x^4+4 x^2}+x^2}$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x^2}{x^2}}{\frac{\sqrt{x^4+4 x^2}+x^2}{x^2}} \cdots\left[\begin{array}{l}
\text { Divide numerator and } \\ \text { denominator by } x^2 \end{array}\right]$
$=\lim _{x \rightarrow \infty} \frac{\frac{4 x^2}{x^2}}{\sqrt{\frac{x^4+4 x^2}{x^4}+1}}$
$=\frac{\lim _{x \rightarrow \infty} 4}{\lim _{x \rightarrow \infty}\left(\sqrt{1+\frac{4}{x^2}}+1\right)}$
$=\frac{4}{\sqrt{1+0}+1} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^{ k }}=0, k >0\right]$
$=\frac{4}{2}$
$=2$
View full question & answer→Question 133 Marks
Evaluate the following limits: $\lim _{x \rightarrow \infty}\left[\frac{a x^3+b x^2+c x+d}{e x^3+f x^2+g x+h}\right]$
Answer$\lim _{x \rightarrow \infty}\left[\frac{a x^3+b x^2+c x+d}{e x^3+f x^2+g x+h}\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{\frac{a x^3+b x^2+c x+d}{x^3}}{\frac{e x^3+f x^2+g x+h}{x^3}}\right]$
$... \left[\begin{array}{l}\text { Divide numerator and } \\ \text { denominator by } x^3\end{array}\right]$
$=\frac{\lim _{x \rightarrow \infty}\left(\mathrm{a}+\frac{\mathrm{b}}{x}+\frac{\mathrm{c}}{x^2}+\frac{\mathrm{d}}{x^3}\right)}{\lim _{x \rightarrow \infty}\left(\mathrm{e}+\frac{\mathrm{f}}{x}+\frac{\mathrm{g}}{x^2}+\frac{\mathrm{h}}{x^3}\right)}$
$=\frac{\lim _{x \rightarrow \infty} \mathrm{a}+\lim _{x \rightarrow \infty} \frac{\mathrm{b}}{x}+\lim _{x \rightarrow \infty} \frac{\mathrm{c}}{x^2}+\lim _{x \rightarrow \infty} \frac{\mathrm{d}}{x^3}}{\lim _{x \rightarrow \infty} \mathrm{e}+\lim _{x \rightarrow \infty} \frac{\mathrm{f}}{x}+\lim _{x \rightarrow \infty} \frac{\mathrm{g}}{x^2}+\lim _{x \rightarrow \infty} \frac{\mathrm{h}}{x^3}}$
$=\frac{\mathrm{a}+0+0+0}{\mathrm{e}+0+0+0} \cdots\left[\lim _{x \rightarrow \infty} \frac{1}{x^{\mathrm{k}}}=0, \mathrm{k}>0\right]$
$=\frac{\mathrm{a}}{\mathrm{e}}$
View full question & answer→Question 143 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{15^x-5^x-3^x+1}{x \cdot \sin x}\right]$
Answer$ \lim _{x \rightarrow 0} \frac{15^x-5^x-3^x+1}{x \sin x}=\lim _{x \rightarrow 0} \frac{5^x \cdot 3^x-5^x-3^x+1}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{5^x\left(3^x-1\right)-1\left(3^x-1\right)}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)\left(5^x-1\right)}{x \sin x}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(3^x-1\right)\left(5^x-1\right)}{x^2}}{\frac{(x \sin x)}{x^2}}$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{3^x-1}{x}\right)\left(\frac{5^x-1}{x}\right)}{\left(\frac{\sin x}{x}\right)}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)}$
$=\frac{(\log 3) \cdot(\log 5)}{1}$
$\ldots\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a, \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=(\log 3) \cdot(\log 5)$
View full question & answer→Question 153 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{\left(2^x-1\right)^3}{\left(3^x-1\right) \cdot \sin x \cdot \log (1+x)}\right]$
Answer$\lim _{x \rightarrow 0} \frac{\left(2^x-1\right)^3}{\left(3^x-1\right) \cdot \sin x \cdot \log (1+x)}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(2^x-1\right)^3}{x^3}}{\frac{\left(3^x-1\right) \cdot \sin x \cdot \log (1+x)}{x^3}}$
${\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x^3 . \\ \because x \rightarrow 0, x \neq 0 \\ \therefore x^3 \neq 0 \end{array}\right]}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)^3}{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \frac{\sin x}{x} \cdot \frac{\log (1+x)}{x}}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{2^x-1}{x}\right)^3}{\left(\lim _{x \rightarrow 0} \frac{3^x-1}{x}\right) \cdot\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)\left(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}\right)}$
$=\frac{(\log 2)^3}{(\log 3)(1)(1)}$
$\ldots\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{x \rightarrow 0} \frac{\mathbf{a}^x-1}{x}=\log a, \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]$
$=\frac{(\log 2)^3}{\log 3}$
View full question & answer→Question 163 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{a^x-b^x}{\sin (4 x)-\sin (2 x)}\right]$
Answer$\lim _{x \rightarrow 0} \frac{a^x-b^x}{\sin 4 x-\sin 2 x}$
$=\lim _{x \rightarrow 0} \frac{\left(\mathrm{a}^x-1\right)-\left(\mathrm{b}^x-1\right)}{2 \cos \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)}$
$=\lim _{x \rightarrow 0} \frac{\left(\mathrm{a}^x-1\right)-\left(\mathrm{b}^x-1\right)}{2 \cos 3 x \cdot \sin x}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(\mathrm{a}^x-1\right)-\left(\mathrm{b}^x-1\right)}{x}}{\frac{2 \cos 3 x \cdot \sin x}{x}} \cdots\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x . \\ \because x \rightarrow 0, x \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{a^x-1}{x}-\frac{b^x-1}{x}\right)}{\lim _{x \rightarrow 0} 2 \cos 3 x \cdot\left(\frac{\sin x}{x}\right)}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}\right)-\left(\lim _{x \rightarrow 0} \frac{\mathrm{b}^x-1}{x}\right)}{2 \lim _{x \rightarrow 0}(\cos 3 x) \cdot \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)}$
$=\frac{\log a-\log b}{2(1)(1)}$
$\ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right]$
$=\frac{1}{2} \log \left(\frac{a}{b}\right)$
View full question & answer→Question 173 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{3+x}{3-x}\right]^{\frac{1}{x}}$
Answer$\lim _{x \rightarrow 0}\left(\frac{3+x}{3-x}\right)^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\frac{x}{3}}{1-\frac{x}{3}}\right)^{\frac{1}{x}} \quad \cdots\left[\begin{array}{l}
\text { Divide numerator and } \\ \text { denominator by } 3 \end{array}\right]$
$=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{3}\right)^{\frac{1}{x}}}{\left(1-\frac{x}{3}\right)^{\frac{1}{x}}}$
$=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{3}\right)^{\frac{3}{x} \times \frac{1}{3}}}{\left(1-\frac{x}{3}\right)^{\frac{-3}{x} \times \frac{1}{3}}}$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{x}{3}\right)^{\frac{3}{x}}\right]^{\frac{1}{3}}}{\lim _{x \rightarrow 0}\left[\left(1-\frac{x}{3}\right)^{\frac{-3}{x}}\right]^{-\frac{1}{3}}}$
$=\frac{\mathrm{e}^{y / 3}}{\mathrm{e}^{-\sqrt{3}}} \quad \ldots\left[\begin{array}{l}
\because x \rightarrow 0, \frac{x}{3} \rightarrow 0, \frac{-x}{3} \rightarrow 0 \text { and } \\
\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=\mathrm{e} \end{array}\right]$
$=\mathrm{e}^{\frac{1}{3}+\frac{1}{3}}$
$=\mathrm{e}^{\frac{2}{3}}$
View full question & answer→Question 183 Marks
Evaluate the following limits: $\lim \limits{x \rightarrow 0}\left[\frac{3^x+3^{-x}-2}{x \cdot \tan x}\right]$
Answer$\lim _{x \rightarrow 0} \frac{3^x+3^{-x}-2}{x \cdot \tan x}$
$=\lim _{x \rightarrow 0} \frac{3^x+\frac{1}{3^x}-2}{x \cdot \tan x}$
$=\lim _{x \rightarrow 0} \frac{3^x+\frac{1}{3^x}-2}{x \cdot \tan x}$
$=\lim _{x \rightarrow 0} \frac{\left(3^x\right)^2+1-2\left(3^x\right)}{3^x \cdot x \tan x}$
$=\lim _{x \rightarrow 0} \frac{\left(3^x-1\right)^2}{3^x \cdot x \cdot \tan x} \cdots\left[\because a^2-2 a b+b^2=(a-b)^2\right]$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(3^x-1\right)^2}{x^2}}{\frac{3^x \cdot x \cdot \tan x}{x^2}} \quad \cdots\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x^2 . \\ \because x \rightarrow 0, x \neq 0 \\ \therefore x^2 \neq 0 \end{array}\right] \\
=\frac{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right)^2}{\lim _{x \rightarrow 0} 3^x \cdot\left(\frac{\tan x}{x}\right)}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{3^x-1}{x}\right)^2}{\left(\lim _{x \rightarrow 0} 3^x\right)\left(\lim _{x \rightarrow 0} \frac{\tan x}{x}\right)}$
$=\frac{(\log 3)^2}{3^0 \times 1} \cdots\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}, \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\right]$
$=(\log 3)^2$
View full question & answer→Question 193 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left(\frac{6^x+5^x+4^x-3^{x+1}}{\sin x}\right)$
Answer$\lim _{x \rightarrow 0} \frac{6^x+5^x+4^x-3^{x+1}}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{6^x+5^x+4^x-3^x \cdot 3^1}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{6^x-1+5^x-1+4^x-1-3^x(3)+3}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(6^x-1\right)+\left(5^x-1\right)+\left(4^x-1\right)-3\left(3^x-1\right)}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(6^x-1\right)+\left(5^x-1\right)+\left(4^x-1\right)-3\left(3^x-1\right)}{\frac{x}{x}}
\ldots\left[\begin{array}{l} \text { Divide Numerator and } \\ \text { Denominator by } x . \\ \because x \rightarrow 0, x \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left[\frac{6^x-1}{x}+\frac{5^x-1}{x}+\frac{4^x-1}{x}-3\left(\frac{3^x-1}{x}\right)\right]}{\lim _{x \rightarrow 0} \frac{\sin x}{x}}$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{6^x-1}{x}\right)+\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)+\lim _{x \rightarrow 0}\left(\frac{4^x-1}{x}\right)-3 \lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right)}{\lim _{x \rightarrow 0} \frac{\sin x}{x}}$
$=\frac{\log 6+\log 5+\log 4-3 \log 3}{1} \ldots\left[\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log \mathrm{a}\right]$
$=\log (6 \times 5 \times 4)-\log (3)^3$
$=\log \frac{120}{27}$
$=\log \frac{40}{9}$
View full question & answer→Question 203 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{9^x-5^x}{4^x-1}\right]$
Answer$\lim _{x \rightarrow 0} \frac{9^x-5^x}{4^x-1} =\lim _{x \rightarrow 0} \frac{9^x-1+1-5^x}{4^x-1}$
$ =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)-\left(5^x-1\right)}{4^x-1}$
$ =\lim _{x \rightarrow 0} \frac{\frac{\left(9^x-1\right)-\left(5^x-1\right)}{x}}{\frac{\left(4^x-1\right)}{x}} $
$[\because x \rightarrow 0, x \neq 0]$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{9^x-1}{x}\right)-\left(\frac{5^x-1}{x}\right)}{\left(\frac{4^x-1}{x}\right)}$
$=\frac{\lim _{x \rightarrow 0} \frac{9^x-1}{x}-\lim _{x \rightarrow 0} \frac{5^x-1}{x}}{\lim _{x \rightarrow 0} \frac{4^x-1}{x}}$
$=\frac{\log 9-\log 5}{\log 4} \ldots\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right]$
$=\frac{1}{(\log 4)} \log \left(\frac{9}{5}\right) $
View full question & answer→Question 213 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 1}\left[\frac{1-x^2}{\sin \pi x}\right]$
Answer$\lim _{x \rightarrow 1} \frac{1-x^2}{\sin \pi x}=\lim _{x \rightarrow 1} \frac{(1-x)(1+x)}{\sin \pi x}$
Put $1-x=\mathrm{h}$,
$\therefore \quad x=1-\mathrm{h}$
$\text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$
\begin{aligned}
\therefore \quad & \lim _{x \rightarrow 1} \frac{(1-x)(1+x)}{\sin \pi x} \\
= & \lim _{h \rightarrow 0} \frac{(h)(1+1-h)}{\sin \pi(1-h)} \\
= & \lim _{h \rightarrow 0} \frac{(h)(2-h)}{\sin (\pi-\pi h)} \\
& =\lim _{h \rightarrow 0} \frac{(h)(2-h)}{\sin \pi h} \quad \ldots[\because \sin (\pi-\theta)=\sin \theta]
\end{aligned}
$
$ =\lim _{h \rightarrow 0} \frac{(2-h)^{}}{\left(\frac{\sin \pi h}{\pi h}\right) \cdot \pi}$
$=\frac{1}{\pi} \cdot \frac{\lim _{h \rightarrow 0}(2-h)}{\lim _{h \rightarrow 0}\left(\frac{\sin \pi h}{\pi h}\right)}$
$=\frac{1}{\pi} \cdot \frac{(2-0)}{1} \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1\right] $
$=\frac{2}{\pi}$
View full question & answer→Question 223 Marks
Evaluate the following limits: $\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{\tan ^2 x-\cot ^2 x}{\sec x-\operatorname{cosec} x}\right]$
Answer$ \lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan ^2 x-\cot ^2 x}{\sec x-\operatorname{cosec} x}$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\left(\sec ^2 x-1\right)-\left(\operatorname{cosec}^2 x-1\right)}{\sec x-\operatorname{cosec} x}$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sec ^2 x-\operatorname{cosec}^2 x}{\sec x-\operatorname{cosec} x}$
$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{(\sec x-\operatorname{cosec} x)(\sec x+\operatorname{cosec} x)}{(\sec x-\operatorname{cosec} x)}$
$=\lim _{x \rightarrow \frac{\pi}{4}}(\sec x+\operatorname{cosec} x)\cdots\left[\begin{array}{l}
\because x \rightarrow \frac{\pi}{4}, \sec x \rightarrow \sqrt{2} \text { and } \operatorname{cosec} x \rightarrow \sqrt{2} \\
\therefore \sec x-\operatorname{cosec} x \rightarrow 0, \therefore \sec x-\operatorname{cosec} x \neq 0
\end{array}\right]$
$=\lim _{x \rightarrow \frac{\pi}{4}}(\sec x)+\lim _{x \rightarrow \frac{\pi}{4}}(\operatorname{cosec} x)$
$=\sec \frac{\pi}{4}+\operatorname{cosec} \frac{\pi}{4}$
$=\sqrt{2}+\sqrt{2}=2 \sqrt{2} $
View full question & answer→Question 233 Marks
Evaluate the following limits:$
\lim _{x \rightarrow \frac{x}{4}}\left[\frac{\cos x-\sin x}{\cos 2 x}\right]
$
Answer$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos 2 x} \\
= & \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{\cos ^2 x-\sin ^2 x} \\
= & \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos x-\sin x}{(\cos x+\sin x)(\cos x-\sin x)} \\
= & \lim _{x \rightarrow \frac{\pi}{4}} \frac{1}{\cos x+\sin x} \\
= & \frac{1}{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}} \\
= & \left.\frac{1}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}=\frac{1}{\frac{2}{\sqrt{2}}}=\frac{\cos x-\sin x \rightarrow 0, \therefore \cos x-\sin x \neq 0}{\sqrt{2}} \text { and } \sin x \rightarrow \frac{1}{\sqrt{2}}\right]
\end{aligned}
$
View full question & answer→Question 243 Marks
Evaluate the following limits:$
\lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-\operatorname{cosec} x}{\cot ^2 x-3}\right]
$
Answer$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{6}}\left[\frac{2-\operatorname{cosec} x}{\cot ^2 x-3}\right]=\lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\operatorname{cosec} x}{\operatorname{cosec}^2 x-1-3} \\
= & \lim _{x \rightarrow \frac{\pi}{6}} \frac{2-\operatorname{cosec} x}{\operatorname{cosec}^2 x-4} \\
= & \lim _{x \rightarrow \frac{\pi}{6}} \frac{-(\operatorname{cosec} x-2)}{(\operatorname{cosec} x-2)(\operatorname{cosec} x+2)} \\
= & \lim _{x \rightarrow \frac{\pi}{6}} \frac{-1}{(\operatorname{cosec} x+2)} \\
= & \frac{-1}{\operatorname{cosec}\left(\frac{\pi}{6}\right)+2}\\
= & \frac{-1}{2+2}=\frac{-1}{4}
\end{aligned}
$
View full question & answer→Question 253 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{x \cdot \tan x}{1-\cos x}\right]$
Answer$\lim _{x \rightarrow 0} \frac{x \cdot \tan x}{1-\cos x}$
$=\lim _{x \rightarrow 0}\left(\frac{x \cdot \tan x}{1-\cos x} \times \frac{1+\cos x}{1+\cos x}\right)$
$=\lim _{x \rightarrow 0} \frac{x \cdot \tan x(1+\cos x)}{1-\cos ^2 x}$
$=\lim _{x \rightarrow 0} \frac{x \cdot \tan x(1+\cos x)}{\sin ^2 x}$
$=\lim _{x \rightarrow 0} \frac{\frac{x \cdot \tan x}{x^2}(1+\cos x)}{\frac{\sin ^2 x}{x^2}}
\cdots\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x^2 \\ \because x \rightarrow 0, x \neq 0 \therefore x^2 \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left[\frac{\tan x}{x}(1+\cos x)\right]}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2}$
$=\frac{\lim _{x \rightarrow 0} \frac{\tan x}{x} \times \lim _{x \rightarrow 0}(1+\cos x)}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2}$
$=\frac{1 \times(1+1)}{(1)^2} \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1, \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\right]$
$=2$
View full question & answer→Question 263 Marks
Evaluate the following limits: $\lim _{\theta \rightarrow 0}\left[\frac{\sin (m \theta)}{\tan (n \theta)}\right]$
Answer$\lim _{\theta \rightarrow 0}\left[\frac{\sin (m \theta)}{\tan (n \theta)}\right]$
$=\lim _{\theta \rightarrow 0}\left[\frac{\frac{\sin (m \theta)}{\theta}}{\frac{\tan (n \theta)}{\theta}}\right]
\cdots\left[\begin{array}{l}
\text { Divide Numerator and Denominator by } \theta \\
\because \theta \rightarrow 0, \quad \theta \neq 0
\end{array}\right]$
$=\frac{\lim _{\theta \rightarrow 0}\left[\frac{\sin (\mathrm{m} \theta)}{\theta}\right]}{\lim _{\theta \rightarrow 0}\left[\frac{\tan (n \theta)}{\theta}\right]}$
$=\frac{\lim _{\theta \rightarrow 0}\left[\frac{\sin (m \theta)}{m \theta} \times m\right]}{\lim _{\theta \rightarrow 0}\left[\frac{\tan (n \theta)}{n \theta} \times n\right]}$
$=\frac{m}{n} \frac{\lim _{\theta \rightarrow 0} \frac{\sin (m \theta)}{m \theta}}{\lim _{\theta \rightarrow 0} \frac{\tan (n \theta)}{n \theta}}$
$=\frac{m}{n} \frac{(1)}{(1)} \cdots\left[\because \lim _{\theta \rightarrow 0} \frac{\sin p \theta}{p \theta}=1, \lim _{\theta \rightarrow 0} \frac{\tan p \theta}{p \theta}=1\right]$
$=\frac{m}{n}$
View full question & answer→Question 273 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right)$
Answer$ \lim _{x \rightarrow 0}\left(\frac{3}{x \sqrt{9-x}}-\frac{1}{x}\right)$
$=\lim _{x \rightarrow 0}\left[\frac{3-\sqrt{9-x}}{x \sqrt{9-x}}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{3-\sqrt{9-x}}{x \sqrt{9-x}} \times \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}}\right]$
$\text {...[By rationalization] }$
$=\lim _{x \rightarrow 0}\left[\frac{9-(9-x)}{x \sqrt{9-x}(3+\sqrt{9-x})}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{x}{x \sqrt{9-x}(3+\sqrt{9-x})}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{1}{\sqrt{9-x}(3+\sqrt{9-x})}\right] \ldots[\because x \rightarrow 0, x \neq 0]$
$=\frac{1}{\sqrt{9-0}(3+\sqrt{9-0})}$
$=\frac{1}{3(3+3)}$
$=\frac{1}{18}$
View full question & answer→Question 283 Marks
Evaluate the following limits: $\lim _{x \rightarrow 4}\left[\frac{x^2+x-20}{\sqrt{3 x+4}-4}\right]$
Answer$\lim _{x \rightarrow 4}\left[\frac{x^2+x-20}{\sqrt{3 x+4}-4}\right]$
$=\lim _{x \rightarrow 4}\left[\frac{(x+5)(x-4)}{\sqrt{3 x+4}-4} \times \frac{\sqrt{3 x+4}+4}{\sqrt{3 x+4}+4}\right]$
$=\lim _{x \rightarrow 4} \frac{(x+5)(x-4)(\sqrt{3 x+4}+4)}{(3 x+4)-16}$
$=\lim _{x \rightarrow 4} \frac{(x+5)(x-4)(\sqrt{3 x+4}+4)}{3 x-12}$
$=\lim _{x \rightarrow 4} \frac{(x+5)(x-4)(\sqrt{3 x+4}+4)}{3(x-4)}$
$=\lim _{x \rightarrow 4} \frac{(x+5)(\sqrt{3 x+4}+4)}{3} \quad \cdots\left[\begin{array}{l}
\because x \rightarrow 4, x \neq 4 \\ \therefore x-4 \neq 0 \end{array}\right]$
$=\frac{(4+5)(\sqrt{3(4)+4}+4)}{3}$
$=\frac{9(4+4)}{3}$
$=24...[$ By rationalization$]$
View full question & answer→Question 293 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 1}\left[\frac{x^2+x \sqrt{x}-2}{x-1}\right]$
Answer$\lim _{x \rightarrow 1}\left[\frac{x^2+x \sqrt{x}-2}{x-1}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{\left(x^2-1\right)+(x \sqrt{x}-1)}{x-1}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{x^2-1}{x-1}+\frac{x^{\frac{3}{2}}-1}{x-1}\right] \ldots\left[\because x \sqrt{x}=x^1 \cdot x^{\frac{1}{2}}=x^{1+\frac{1}{2}}=x^{\frac{3}{2}}\right]$
$=\lim _{x \rightarrow 1}\left(\frac{x^2-1^2}{x-1}\right)+\lim _{x \rightarrow 1}\left(\frac{x^{\frac{3}{2}}-1^{\frac{3}{2}}}{x-1}\right) \quad \ldots\left[\because \lim _{x \rightarrow 2} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\right.\text { na }\left.=2(1)^1+\frac{3}{2}(1)^{\frac{1}{2}}\right]$
$=2+\frac{3}{2}=\frac{7}{2}$
View full question & answer→Question 303 Marks
Evaluate the following limits: $\lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]$
Answer$ \lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right]$
$=\lim _{x \rightarrow 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}} \times \frac{\sqrt{x+2}+\sqrt{3 x-2}}{\sqrt{x+2}+\sqrt{3 x-2}}\right]......[$ By rationalization $]$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{x+2}+\sqrt{3 x-2})}{(x+2)-(3 x-2)}$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(\sqrt{x+2}+\sqrt{3 x-2})}{-2 x+4}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)(\sqrt{x+2}+\sqrt{3 x-2})}{-2(x-2)}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(\sqrt{x+2}+\sqrt{3 x-2})}{-2}$
$=\frac{(2+2)\left(\because x \rightarrow 2, x \neq 2$
$\therefore x-2 \neq 0\right]}{-2}$
$=\frac{4(2+2)}{-2}$
$=\frac{16}{-2}$
$=-8 $
View full question & answer→Question 313 Marks
Evaluate the following limits: $\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^2-9}\right]$
Answer$\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^2-9}\right]$
$= \lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+3}-\sqrt{4 x-3}}{x^2-9} \times \frac{\sqrt{2 x+3}+\sqrt{4 x-3}}{\sqrt{2 x+3}+\sqrt{4 x-3}}\right]......[$By rationalization$]$
$=\lim _{x \rightarrow 3}\left[\frac{(2 x+3)-(4 x-3)}{\left(x^2-9\right)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{-2 x+6}{\left(x^2-9\right)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{-2(x-3)}{(x+3)(x-3)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{-2}{(x+3)(\sqrt{2 x+3}+\sqrt{4 x-3})}\right]$
$\text {}[x x \rightarrow 3 \neq 0 ; \neq 0$
$=\frac{-2}{(3+3)(\sqrt{2(3)+3}+\sqrt{4(3)-3})}$
$=\frac{-2}{6(3+3)}$
$=\frac{-1}{18}$
View full question & answer→Question 323 Marks
Evaluate the following limits: $\lim _{x \rightarrow a}\left[\frac{1}{x^2-3 a x+2 a^2}+\frac{1}{2 x^2-3 a x+a^2}\right]$
Answer$
\lim _{x \rightarrow \mathrm{a}}\left[\frac{1}{x^2-3 \mathrm{a} x+2 \mathrm{a}^2}+\frac{1}{2 x^2-3 \mathrm{a} x+\mathrm{a}^2}\right]
$
Consider,
$x^2-3 \mathrm{ax}+2 \mathrm{a}^2 =x^2-2 \mathrm{ax}-\mathrm{ax}+2 \mathrm{a}^2$
$ =x(x-2 \mathrm{a})-\mathrm{a}(x-2 \mathrm{a})$
$ =(x-2 \mathrm{a})(x-\mathrm{a})$
$2 x^2-3 \mathrm{ax}+\mathrm{a}^2 =2 x^2-2 \mathrm{a} x-\mathrm{a} x+\mathrm{a}^2$
$ =2 x(x-\mathrm{a})-\mathrm{a}(x-\mathrm{a})$
$ =(x-a)(2 x-a)$
$\therefore \lim _{x \rightarrow 0}\left[\frac{1}{x^2-3 \mathrm{a} x+2 \mathrm{a}^2}+\frac{1}{2 x^2-3 \mathrm{a} x+\mathrm{a}^2}\right]$
$ =\lim _{x \rightarrow \mathrm{a}}\left[\frac{1}{(x-2 \mathrm{a})(x-\mathrm{a})}+\frac{1}{(x-\mathrm{a})(2 x-\mathrm{a})}\right]$
$ =\lim _{x \rightarrow \mathrm{a}} \frac{(2 x-\mathrm{a})+(x-2 \mathrm{a})}{(x-2 \mathrm{a})(x-\mathrm{a})(2 x-\mathrm{a})}$
$ =\lim _{x \rightarrow \mathrm{a}} \frac{3 x-3 \mathrm{a}}{(x-2 \mathrm{a})(x-\mathrm{a})(2 x-\mathrm{a})}$
$ =\lim _{x \rightarrow \mathrm{a}} \frac{3(x-\mathrm{a})}{(x-2 \mathrm{a})(x-\mathrm{a})(2 x-\mathrm{a})} \quad\left[\begin{array}{l} \because x \rightarrow \mathrm{a}, x \neq \mathrm{a} \\ \therefore x-\mathrm{a} \neq 0 \end{array}\right]$
$=\lim _{x \rightarrow \mathrm{a}} \frac{3}{(x-2 \mathrm{a})(2 x-\mathrm{a})}$
$ =\frac{3}{(\mathrm{a}-2 \mathrm{a})(2 \mathrm{a}-\mathrm{a})}$
$ =\frac{3}{(-\mathrm{a})(\mathrm{a})}=\frac{-3}{\mathrm{a}^2}$
View full question & answer→Question 333 Marks
Evaluate the following limits: $\lim _{x \rightarrow 1}\left[\frac{x+2}{x^2-5 x+4}+\frac{x-4}{3\left(x^2-3 x+2\right)}\right]$
View full question & answer→Question 343 Marks
Evaluate the following limits: $\lim _{x \rightarrow 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3 x^2+2 x}\right]$
Answer$\lim _{x \rightarrow 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3 x^2+2 x}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x\left(x^2-3 x+2\right)}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{x-2}{x(x-1)}-\frac{1}{x(x-2)(x-1)}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{(x-2)^2-1}{x(x-2)(x-1)}\right]$
$=\lim _{x \rightarrow 1} \frac{x^2-4 x+4-1}{x(x-2)(x-1)}$
$=\lim _{x \rightarrow 1} \frac{x^2-4 x+3}{x(x-2)(x-1)}$
$=\lim _{x \rightarrow 1} \frac{(x-3)(x-1)}{x(x-2)(x-1)}$
$=\lim _{x \rightarrow 1} \frac{x-3}{x(x-2)} \cdots\left[\begin{array}{l} \because x \rightarrow 1, x \neq 1, \\ \therefore x-1 \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 1}(x-3)}{\lim _{x \rightarrow 1} x(x-2)}$
$=\frac{1-3}{1(1-2)}$
$=\frac{-2}{-1}$
$=2$
View full question & answer→Question 353 Marks
Evaluate the following limits: $\lim _{y \rightarrow \frac{1}{2}}\left[\frac{1-8 y^3}{y-4 y^3}\right]$
Answer$\lim _{y \rightarrow \frac{1}{2}}\left[\frac{1-8 y^3}{y-4 y^3}\right]$
$=\lim _{y \rightarrow \frac{1}{2}} \frac{1-8 y^3}{y\left(1-4 y^2\right)}$
$=\lim _{y \rightarrow \frac{1}{2}} \frac{(1)^3-(2 y)^3}{y\left[(1)^2-(2 y)^2\right]}$
$=\lim _{y \rightarrow \frac{1}{2}} \frac{(1-2 y)\left(1+2 y+4 y^2\right)}{y(1-2 y)(1+2 y)}$
$=\lim _{y \rightarrow \frac{1}{2}} \frac{1+2 y+4 y^2}{y(1+2 y)} \quad \cdots\left[\begin{array}{l} \because y \rightarrow \frac{1}{2}, y \neq \frac{1}{2} \\ \therefore 2 y \neq 1, \therefore 2 y-1 \neq 0 \\ \therefore 1-2 y \neq 0 \end{array}\right]$
$=\frac{\lim _{y \rightarrow \frac{1}{2}}\left(1+2 y+4 y^2\right)}{\lim _{y \rightarrow \frac{1}{2}}[y(1+2 y)]}$
$=\frac{1+2\left(\frac{1}{2}\right)+4\left(\frac{1}{2}\right)^2}{\frac{1}{2}\left[1+2\left(\frac{1}{2}\right)\right]}$
$=\frac{1+1+1}{\frac{1}{2}(2)}$
$=3$
View full question & answer→Question 363 Marks
Evaluate the following limits: $\lim _{x \rightarrow \sqrt{2}}\left[\frac{x^2+x \sqrt{2}-4}{x^2-3 x \sqrt{2}+4}\right]$
Answer$\lim _{x \rightarrow \sqrt{2}}\left[\frac{x^2+x \sqrt{2}-4}{x^2-3 x \sqrt{2}+4}\right]$
Consider, $x^2+x \sqrt{2}-4=x^2+\sqrt{2} x-4$
$=x^2+2 \sqrt{2} x-\sqrt{2} x-4$
$=x(x+2 \sqrt{2})-\sqrt{2}(x+2 \sqrt{2})$
$=(x+2 \sqrt{2})(x-\sqrt{2})$
$x^2-3 x \sqrt{2}+4 =x^2-3 \sqrt{2} x+4$
$= x^2-2 \sqrt{2} x-\sqrt{2} x+4$
$= x(x-2 \sqrt{2})-\sqrt{2}(x-2 \sqrt{2})$
$= (x-2 \sqrt{2})(x-\sqrt{2})$
Now, $\lim _{x \rightarrow \sqrt{2}}\left[\frac{x^2+x \sqrt{2}-4}{x^2-3 x \sqrt{2}+4}\right]$
$=\lim _{x \rightarrow \sqrt{2}} \frac{(x+2 \sqrt{2})(x-\sqrt{2})}{(x-2 \sqrt{2})(x-\sqrt{2})}$
$=\lim _{x \rightarrow \sqrt{2}} \frac{x+2 \sqrt{2}}{x-2 \sqrt{2}} \quad \ldots\left[\begin{array}{l}
\because x \rightarrow \sqrt{2}, x \neq \sqrt{2}, \\ \therefore x-\sqrt{2} \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow \sqrt{2}}(x+2 \sqrt{2})}{\lim _{x \rightarrow \sqrt{2}}(x-2 \sqrt{2})}$
$=\frac{\sqrt{2}+2 \sqrt{2}}{\sqrt{2}-2 \sqrt{2}}$
$=\frac{3 \sqrt{2}}{-\sqrt{2}}$
$=-3$
View full question & answer→Question 373 Marks
Evaluate the following limits: $\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^3-27}\right]$
Answer$\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^3-27}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^3-3^3}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{(x-3)\left(x^2+3 x+9\right)}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{x^2+3 x+9-9 x}{(x-3)\left(x^2+3 x+9\right)}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{x^2-6 x+9}{(x-3)\left(x^2+3 x+9\right)}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{(x-3)^2}{(x-3)\left(x^2+3 x+9\right)}\right]$
$=\lim _{x \rightarrow 3}\left[\frac{x-3}{x^2+3 x+9}\right] \quad \cdots\left[\begin{array}{l}\because x \rightarrow 3, x \neq 3, \\\therefore x-3 \neq 0\end{array}\right] \\
=\frac{\lim _{x \rightarrow 3}(x-3)}{\lim _{x \rightarrow 3}\left(x^2+3 x+9\right)}$
$=\frac{3-3}{(3)^2+3(3)+9}$
$=\frac{0}{3}$
$=0$
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Evaluate the following limits:$\lim _{x \rightarrow 1}\left(\frac{x+x^3+x^5+\ldots+x^{2 n-1}-n}{x-1}\right)$
Answer$
\begin{aligned}
& \lim _{x \rightarrow 1}\left[\frac{x+x^3+x^5+\ldots+x^{2 n-1}-\mathrm{n}}{x-1}\right] \\
= & \lim _{x \rightarrow 1}\left[\frac{x+x^3+x^5+\ldots+x^{2 n-1}-(1+1+1+\ldots \mathrm{n} \text { times })}{x-1}\right] \\
= & \lim _{x \rightarrow 1}\left[\frac{(x-1)+\left(x^3-1\right)+\left(x^5-1\right)+\ldots+\left(x^{2 n-1}-1\right) \ldots(\text { n brackets })}{x-1}\right] \\
& \ldots \lim _{x \rightarrow 1}\left[\frac{x^1-1^1}{x-1}+\frac{x^3-1^3}{x-1}+\frac{x^5-1^5}{x-1}+\ldots+\frac{x^{2 n-1}-1^{2 \mathrm{n}-1}}{x-1}\right] \\
= & 1(1)^0+3(1)^2+5(1)^4+\ldots+(2 \mathrm{n}-1)(1)^{2 \mathrm{n}-2} \\
= & 1+3+5+\ldots+(2 \mathrm{n}-1) \\
= & \sum_{\mathrm{r}=1}^{\mathrm{n}}(2 \mathrm{r}-1) \\
= & 2 \sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}-\sum_{\mathrm{r}=1}^{\mathrm{n}} 1 \\
= & 2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\mathrm{n} \\
= & \mathrm{n}(\mathrm{n}+1)-\mathrm{n} \\
= & \mathrm{n}^2+\mathrm{n}-\mathrm{n} \\
= & \mathrm{n}^2
\end{aligned}
$
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Evaluate the following limits: $\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]$
Answer$\lim _{y \rightarrow 1} \frac{2 y-2}{\sqrt[3]{7+y}-2}$
$=\lim _{y \rightarrow 1} \frac{2(y-1)}{(7+y)^{\frac{1}{3}}-8^{\frac{1}{3}}} \cdots\left[\because 2=\left(2^3\right)^{\frac{1}{3}}=8^{\frac{1}{3}}\right]$
$=\lim _{y \rightarrow 1} \frac{2}{\frac{(y+7)^{\frac{1}{3}}-8^{\frac{1}{3}}}{y-1}}$
$=\frac{\lim _{y \rightarrow 1} 2}{\lim _{y \rightarrow 1} \frac{(y+7)^{\frac{1}{3}}-8^{\frac{1}{3}}}{(y+7)-8}}$
$=\frac{2}{\frac{1}{3}(8)^{\frac{-2}{3}}} \quad \ldots\left[\begin{array}{l} \because y \rightarrow 1, y+7 \rightarrow 8 \\
\text { and } \lim _{x \rightarrow a} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{\mathrm{n}-1}
\end{array}\right]$
$=2(3) \cdot(8)^{\frac{2}{3}}$
$=6\left(2^3\right)^{\frac{2}{3}}$
$=6 \times(2)^2$
$=24$
View full question & answer→Question 403 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}\right]$
Answer$ \lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}$
$=\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{3}}-(1+x)^{\frac{1}{2}}}{x} $
Put $1+x=y$
As $x \rightarrow 0, y \rightarrow 1$
$ \lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{3}}-(1+x)^{\frac{1}{2}}}{x}$
$=\lim _{y \rightarrow 1} \frac{y^{\frac{1}{3}}-y^{\frac{1}{2}}}{y-1}$
$=\lim _{y \rightarrow 1} \frac{\left(y^{\frac{1}{3}}-1\right)-\left(y^{\frac{1}{2}}-1\right)}{y-1}$
$=\lim _{y \rightarrow 1}\left(\frac{y^{\frac{1}{3}}-1}{y-1}-\frac{y^{\frac{1}{2}}-1}{y-1}\right)$
$=\lim _{y \rightarrow 1} \frac{y^{\frac{1}{3}}-1^{\frac{1}{3}}}{y-1}-\lim _{y \rightarrow 1} \frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1}$
$=\frac{1}{3}(1)^{\frac{-2}{3}}-\frac{1}{2}(1)^{\frac{-1}{2}} \quad \cdots\left[\because \lim _{x \rightarrow a} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{\mathrm{n}-1}\right]$
$=\frac{1}{3}-\frac{1}{2}$
$=\frac{2-3}{6}$
$=-\frac{1}{6}$
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