Question 12 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^x+1}{(x-1)^2}\right]$
Answer$ \lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^x+1}{(x-1)^2}\right]$
$\text { put } x-1=\mathbf{h}$
$\therefore \quad x=1+\mathrm{h}$
$\text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\text { Required limit }$
$=\lim _{h \rightarrow 0} \frac{4^h-2^{1+h}+1}{h^2}$
$=\lim _{h \rightarrow 0} \frac{\left(2^2\right)^h-2^1 \cdot 2^h+1}{h^2}$
$=\lim _{h \rightarrow 0} \frac{\left(2^h\right)^2-2\left(2^h\right)+1}{h^2}$
$=\lim _{h \rightarrow 0} \frac{\left(2^h-1\right)^2}{h^2}$
$=\lim _{h \rightarrow 0}\left(\frac{2^h-1}{h}\right)^2$
$=(\log 2)^2 \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right]$
View full question & answer→Question 22 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 2}\left[\frac{\log x-\log 2}{x-2}\right]$
Answer$\lim _{x \rightarrow 2} \frac{\log x-\log 2}{x-2}$
Put $x-2=\mathrm{h}$
$\therefore \quad x=2+\mathrm{h}$
Required limit
$ =\lim _{h \rightarrow 0} \frac{\log (2+h)-\log 2}{h}$
$=\lim _{h \rightarrow 0} \frac{\log \left(\frac{2+h}{2}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{h}{2}\right)}{\frac{h}{2} \times 2}$
$=1 \times \frac{1}{2}$
$=\frac{1}{2} $
View full question & answer→Question 32 Marks
Evaluate the following limits:Given that $7 \mathrm{x} \leq \mathrm{f}(\mathrm{x}) \leq 3 \mathrm{x}^2-6$ for all $x$. Determine the value of $\lim _{x \rightarrow 3} f(x)$
Answer$
\begin{array}{ll}
& 7 x \leq \mathrm{f}(x) \leq 3 x^2-6 \\
& \text { Taking limits as } x \rightarrow 3 \\
& \lim _{x \rightarrow 3}(7 x) \leq \lim _{x \rightarrow 3} \mathrm{f}(x) \leq \lim _{x \rightarrow 3}\left(3 x^2-6\right) \\
\therefore \quad & 7(3) \leq \lim _{x \rightarrow 3} \mathrm{f}(x) \leq 3(3)^2-6 \\
\therefore \quad & 21 \leq \lim _{x \rightarrow 3} \mathrm{f}(x) \leq 21 \\
&
\end{array}
$
Using squeeze theorem,
$
\lim _{x \rightarrow 3} f(x)=21
$
View full question & answer→Question 42 Marks
Evaluate the following limits:Find the limit of the function, if it exists, at $x=1$
$
f(x)=\left\{\begin{array}{lll}
7-4 x & \text { for } & x<1 \\
x^2+2 & \text { for } & x \geq 1
\end{array}\right.
$
Answer$
\begin{aligned}
\mathrm{f}(x)=7 & -4 x ; x<1 \\
= & x^2+2 ; x \geq 1 \\
\lim _{x \rightarrow 1^{-}} \mathrm{f}(x) & =\lim _{x \rightarrow 1}(7-4 x) \\
& =7-4(1) \\
& =3 \\
\lim _{x \rightarrow 1^{+}} \mathrm{f}(x) & =\lim _{x \rightarrow 1}\left(x^2+2\right) \\
& =(1)^2+2 \\
& =3 \\
\therefore \quad \lim _{x \rightarrow 1^{-}} \mathrm{f}(x) & =\lim _{x \rightarrow 1^{+}} \mathrm{f}(x)
\end{aligned}
$
$\therefore \quad \lim _{x \rightarrow 1^{+}} \mathrm{f}(x)$ exists and is equal to 3 .
View full question & answer→Question 52 Marks
Evaluate the following limits:
If $f(r)=\pi r^2$ then find $\lim _{h \rightarrow 0}\left[\frac{f(r+h)-f(r)}{h}\right]$
Answer$ \mathrm{f}(\mathrm{r})=\pi \mathrm{r}^2$
$\mathrm{f}(\mathrm{r}+\mathrm{h})=\pi(\mathrm{r}+\mathrm{h})^2$
$=\pi\left(r^2+2 r h+h^2\right)$
$=\pi \mathrm{r}^2+2 \pi \mathrm{rh}+\pi \mathrm{h}^2$
$\therefore \quad \lim _{h \rightarrow 0} \frac{f(r+h)-f(r)}{h}$
$=\lim _{h \rightarrow 0} \frac{\pi \mathrm{r}^2+2 \pi \mathrm{h}+\pi \mathrm{h}^2-\pi \mathrm{r}^2}{\mathrm{~h}}$
$=\lim _{h \rightarrow 0} \frac{2 \pi h+\pi h^2}{h M}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}(2 \pi \mathrm{r}+\pi \mathrm{h})}{\mathrm{h}}$
$=\lim _{h \rightarrow 0}(2 \pi r+\pi h) \quad \ldots[\because h \rightarrow 0, h \neq 0]$
$=2 \pi \mathrm{r}+\pi(0)$
$=2 \pi r$
View full question & answer→Question 62 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}[x]$ ([] is a greatest integer function.)
Answer$ \lim _{x \rightarrow 0}[x]$
${[x]=-1 \quad ;-1 \leq x<0}$
${[x]=0 \quad ; 0 \leq x<1}$
$\lim _{x \rightarrow 0^{-}}[x]=-1$
$\lim _{x \rightarrow 0^{+}}[x]=0$
$\therefore \quad \lim _{x \rightarrow 0^{-}}[x] \neq \lim _{x \rightarrow 0^{+}}[x]$
$\therefore \quad \lim _{x \rightarrow 0}[x] \text { does not exist. }$
View full question & answer→Question 72 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{(1-x)^5-1}{(1-x)^3-1}\right]$
Answer$\lim _{x \rightarrow 0}\left[\frac{(1-x)^5-1}{(1-x)^3-1}\right]$
Put $1-x=y$
As $x \rightarrow 0, y \rightarrow 1$
$ =\frac{\lim _{y \rightarrow 1} \frac{y^5-1^5}{y-1}}{\lim _{y \rightarrow 1} \frac{y^3-1^3}{y-1}}$
$=\frac{5(1)^4}{3(1)^2}$
$\ldots\left[\lim _{x \rightarrow a} \frac{x^{\mathrm{n}}-\mathrm{a}^{\mathrm{n}}}{x-\mathrm{a}}=\mathrm{na}^{\mathrm{n}-1}\right]$
$=\frac{5}{3}$
View full question & answer→Question 82 Marks
Evaluate the following limits:
$\lim _{x \rightarrow \infty}\left[\frac{7 x^2+5 x-3}{8 x^2-2 x+7}\right]$
Answer$ \lim _{x \rightarrow \infty} \frac{7 x^2+5 x-3}{8 x^2-2 x+7}$
$=\lim _{x \rightarrow \infty} \frac{\left(\frac{7 x^2+5 x-3}{x^2}\right)}{\left(\frac{8 x^2-2 x+7}{x^2}\right)}$
$=\lim _{x \rightarrow \infty} \frac{\left(7+\frac{5}{x}-\frac{3}{x^2}\right)}{\left(8-\frac{2}{x}+\frac{7}{x^2}\right)}$
$=\frac{\lim _{x \rightarrow \infty} 7+\lim _{x \rightarrow \infty} \frac{5}{x^2}-\lim _{x \rightarrow \infty} \frac{3}{x^2}}{\lim _{x \rightarrow \infty} 8-\lim _{x \rightarrow \infty} \frac{2}{x}+\lim _{x \rightarrow \infty} \frac{7}{x^2}}$
$=\frac{7+0-0}{8-0+0} \quad \ldots\left[\because \lim _{x \rightarrow \infty} \frac{1}{x^k}=0, k >0\right]$
$=\frac{7}{8} $
View full question & answer→Question 92 Marks
Evaluate the following limits : $\lim _{x \rightarrow \infty}\left[\frac{7 x^2+2 x-3}{\sqrt{x^4+x+2}}\right]$
Answer$\lim _{x \rightarrow \infty}\left[\frac{7 x^2+2 x-3}{\sqrt{x^4+x+2}}\right]$
$=\lim _{x \rightarrow \infty} \frac{\frac{7 x^2+2 x-3}{x^2}}{\frac{\sqrt{x^4+x+2}}{x^2}} \cdots[$Divide numerator and denominator by $x^2]$
$=\frac{\lim _{x \rightarrow \infty}\left(7+\frac{2}{x}-\frac{3}{x^2}\right)}{\lim _{x \rightarrow \infty} \sqrt{\frac{x^4+x+2}{x^4}}}$
$=\frac{\lim _{x \rightarrow \infty} 7+\lim _{x \rightarrow \infty} \frac{2}{x}-\lim _{x \rightarrow \infty} \frac{3}{x^2}}{\lim _{x \rightarrow \infty} \sqrt{1+\frac{1}{x^3}+\frac{2}{x^4}}} \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^2}=0, \mathrm{k}>0\right]$
$=\frac{7+0-0}{\sqrt{1+0+0}} 7$
$=7$
View full question & answer→Question 102 Marks
Evaluate the following limits:
$\lim _{x \rightarrow \infty}\left[\frac{x^3+3 x+2}{(x+4)(x-6)(x-3)}\right]$
Answer$ \lim _{x \rightarrow \infty}\left[\frac{x^3+3 x+2}{(x+4)(x-6)(x-3)}\right]$
$=\lim _{x \rightarrow \infty} \frac{\frac{x^3+3 x+2}{x^3}}{\frac{(x+4)(x-6)(x-3)}{x^3}}$
$=\frac{\lim _{x \rightarrow \infty}\left(1+\frac{3}{x^2}+\frac{2}{x^3}\right)}{\lim _{x \rightarrow \infty}\left(\frac{x+4}{x}\right)\left(\frac{x-6}{x}\right)\left(\frac{x-3}{x}\right)}$
$=\frac{\lim _{x \rightarrow \infty} 1+\lim _{x \rightarrow \infty} \frac{3}{x^2}+\lim _{x \rightarrow \infty} \frac{2}{x^3}}{\lim _{x \rightarrow \infty}\left(1+\frac{4}{x}\right) \cdot \lim _{x \rightarrow \infty}\left(1-\frac{6}{x}\right) \cdot \lim _{x \rightarrow \infty}\left(1-\frac{3}{x}\right)}$
$=\frac{1+0+0}{(1+0)(1-0)(1-0)} \cdots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, \mathrm{k}>0\right]$
$=1 $
View full question & answer→Question 112 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{(25)^x-2(5)^x+1}{x \cdot \sin x}\right]$
Answer$\lim _{x \rightarrow 0} \frac{(25)^x-2(5)^x+1}{x \cdot \sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(5^x\right)^2-2\left(5^x\right)+1}{x \cdot \sin x}$ $\cdots\left[(25)^x=\left(5^2\right)^x=\left(5^x\right)^2\right]$
$=\lim _{x \rightarrow 0} \frac{\left(5^x-1\right)^2}{x \cdot \sin x}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(5^x-1\right)^2}{x^2}}{\frac{x \cdot \sin x}{x^2}} \quad \cdots\left[\begin{array}{l} \text { Divide numerator and } \\ \text { denominator by } x^2 . \\ \because x \rightarrow 0, x \neq 0 \\ \therefore x^2 \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)^2 }{\lim _{x \rightarrow 0} \frac{\sin x}{x}}$
$=\frac{(\log 5)^2}{1} \quad \cdots\left[\lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}, \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$
$=(\log 5)^2$
View full question & answer→Question 122 Marks
Evaluate the following limits $\lim _{x \rightarrow 0}\left[\frac{5+7 x}{5-3 x}\right]^{\frac{1}{3 x}}$
Answer$\lim _{x \rightarrow 0}\left(\frac{5+7 x}{5-3 x}\right)^{\frac{1}{3 x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\frac{7}{5} x}{1-\frac{3}{5} x}\right)^{\frac{1}{3 x}} \cdots[$ Divide numerator and denominator by $5]$
$=\lim _{x \rightarrow 0} \frac{\left(1+\frac{7 x}{5}\right)^{\frac{1}{3 x}}}{\left(1-\frac{3 x}{5}\right)^{\frac{1}{3 x}}}$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{7 x}{5}\right)^{\frac{5}{7 x}}\right]^{\frac{7}{5} \times \frac{1}{3}}}{\lim _{x \rightarrow 0}\left[\left(1-\frac{3 x}{5}\right)^{\frac{-5}{3 x}}\right]^{\frac{-3}{5} \times \frac{1}{3}}}$
$=\frac{\mathrm{e}^{\frac{7}{15}}}{\mathrm{e}^{\frac{-3}{15}}} \cdots[
\because x \rightarrow 0, \frac{7 x}{5} \rightarrow 0, \frac{-3 x}{5} \rightarrow 0$ and $\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=\mathrm{e}]$
$=e^{\frac{7}{15}+\frac{3}{15}}=e^{\frac{10}{15}}$
$=\mathrm{e}^{\frac{2}{3}}$
View full question & answer→Question 132 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{5 x+3}{3-2 x}\right]^{\frac{2}{x}}$
Answer$\lim _{x \rightarrow 0}\left(\frac{5 x+3}{3-2 x}\right)^{\frac{2}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-2 x}\right)^{\frac{2}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\frac{5}{3} x}{1-\frac{2}{3} x}\right)^{\frac{2}{x}}$
$=\lim _{x \rightarrow 0} \frac{\left(1+\frac{5 x}{3}\right)^{\frac{2}{x}}}{\left(1-\frac{2 x}{3}\right)^{\frac{2}{x}}}$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{5 x}{3}\right)^{\frac{3}{5 x}}\right]^{\frac{5}{3} \times 2}}{\lim _{x \rightarrow 0}\left[\left(1-\frac{2 x}{3}\right)^{\frac{-3}{2 x}}\right]^{-\frac{2}{3} \times 2}}$
$=\frac{\mathrm{e}^{\frac{10}{3}}}{\mathrm{e}^{\frac{-4}{3}}} \cdots[
\because x \rightarrow 0, \frac{5 x}{3} \rightarrow 0, \frac{-2 x}{3} \rightarrow 0$ and $\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=\mathrm{e}
]$
$=e^{\frac{14}{3}}$
View full question & answer→Question 142 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left(\frac{a^x+b^x+c^x-3}{\sin x}\right)$
Answer$\lim _{x \rightarrow 0} \frac{a^x+b^x+c^x-3}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(a^x-1\right)+\left(b^x-1\right)+\left(c^x-1\right)}{\sin x}$
$=\lim _{x \rightarrow 0} \frac{\left(\mathrm{a}^x-1\right)+\left(\mathrm{b}^x-1\right)+\left(\mathrm{c}^x-1\right)}{x} [$ Divide numerator and denominator by $x \quad \because x \rightarrow 0, x \neq 0]$
$=\lim _{x \rightarrow 0} \frac{\left(\frac{a^x-1}{x}\right)+\left(\frac{b^x-1}{x}\right)+\left(\frac{c^x-1}{x}\right)}{\frac{\sin x}{x}}$
$=\frac{\left(\lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}\right)+\left(\lim _{x \rightarrow 0} \frac{\mathrm{b}^x-1}{x}\right)+\left(\lim _{x \rightarrow 0} \frac{\mathrm{c}^x-1}{x}\right)}{\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)}$
$=\frac{\log \mathrm{a}+\log \mathrm{b}+\log \mathrm{c}}{1} \ldots .\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
$=\log (a b c)$
View full question & answer→Question 152 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{5^x+3^x-2^x-1}{x}\right]$
Answer$ \lim _{x \rightarrow 0} \frac{5^x+3^x-2^x-1}{x}$
$=\lim _{x \rightarrow 0} \frac{\left(5^x-1\right)+\left(3^x-2^x\right)}{x}$
$=\lim _{x \rightarrow 0} \frac{\left(5^x-1\right)+\left(3^x-1\right)-\left(2^x-1\right)}{x}$
$=\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}+\frac{3^x-1}{x}-\frac{2^x-1}{x}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{5^x-1}{x}\right)+\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right)-\lim _{x \rightarrow 0}\left(\frac{2^x-1}{x}\right)$
$=\log 5+\log 3-\log 2 \quad \ldots\left[\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\operatorname{loga}\right]$
$=\log \frac{5 \times 3}{2}$
$=\log \frac{15}{2} $
View full question & answer→Question 162 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left(\frac{\sec x-1}{x^2}\right)$
Answer$ \lim _{x \rightarrow 0} \frac{\sec x-1}{x^2}$
$=\lim _{x \rightarrow 0} \frac{(\sec x-1)(\sec x+1)}{x^2(\sec x+1)}$
$=\lim _{x \rightarrow 0} \frac{\sec ^2 x-1}{x^2(\sec x+1)}=\lim _{x \rightarrow 0} \frac{\tan ^2 x}{x^2(\sec x+1)}$
$=\lim _{x \rightarrow 0}\left[\left(\frac{\tan x}{x}\right)^2 \times \frac{1}{\sec x+1}\right]$
$=\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^2 \times \lim _{x \rightarrow 0} \frac{1}{\sec x+1}$
$=(1)^2 \times \frac{1}{1+1} \quad \ldots\left[\because \lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\right]$
$=\frac{1}{2} $
View full question & answer→Question 172 Marks
Evaluate the following limits:
$\lim _{y \rightarrow 0}\left[\frac{\sqrt{a+y}-\sqrt{a}}{y \sqrt{a+y}}\right]$
Answer$ \lim _{y \rightarrow 0}\left[\frac{\sqrt{\mathrm{a}+y}-\sqrt{\mathrm{a}}}{y \sqrt{\mathrm{a}+y}}\right]$
$\lim _{y \rightarrow 0}\left[\frac{\sqrt{\mathrm{a}+y}-\sqrt{\mathrm{a}}}{y \sqrt{\mathrm{a}+y}} \times \frac{\sqrt{\mathrm{a}+y}+\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}+y}+\sqrt{\mathrm{a}}}\right]$
$=\lim _{y \rightarrow 0} \frac{\mathrm{a}+y-\mathrm{a}}{y \sqrt{\mathrm{a}+y}(\sqrt{\mathrm{a}+y}+\sqrt{\mathrm{a}})}$
$=\lim _{y \rightarrow 0} \frac{y}{y \sqrt{\mathrm{a}+y}(\sqrt{\mathrm{a}+y}+\sqrt{\mathrm{a}})}$
$=\lim _{y \rightarrow 0} \frac{1}{\sqrt{\mathrm{a}+y}(\sqrt{\mathrm{a}+y}+\sqrt{\mathrm{a}})}$
$=\frac{1}{\sqrt{\mathrm{a}+0}(\sqrt{\mathrm{a}+0}+\sqrt{\mathrm{a}})}$
$=\frac{1}{\sqrt{\mathrm{a}}(2 \sqrt{\mathrm{a}})} \quad[\because y \rightarrow 0,$
$\therefore y \neq 0 $
..[By rationalization]
View full question & answer→Question 182 Marks
Evaluate the following limits: $\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^2}-\sqrt{1+y^2}}{y^2}\right]$
Answer$\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^2}-\sqrt{1+y^2}}{y^2}\right]$
$=\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^2}-\sqrt{1+y^2}}{y^2} \times \frac{\sqrt{1-y^2}+\sqrt{1+y^2}}{\sqrt{1-y^2}+\sqrt{1+y^2}}\right]$
$...[$ By rationalization$]$
$=\lim _{y \rightarrow 0} \frac{\left(1-y^2\right)-\left(1+y^2\right)}{y^2\left(\sqrt{1-y^2}+\sqrt{1+y^2}\right)}$
$=\lim _{y \rightarrow 0} \frac{1-y^2-1-y^2}{y^2\left(\sqrt{1-y^2}+\sqrt{1+y^2}\right)}$
$=\lim _{y \rightarrow 0} \frac{-2 y^2}{y^2\left(\sqrt{1-y^2}+\sqrt{1+y^2}\right)}$
$=\lim _{y \rightarrow 0} \frac{-2}{\sqrt{1-y^2}+\sqrt{1+y^2}}$
$\ldots [\because y \rightarrow 0, y \neq 0, \therefore y^2 \neq 0]$
$=\frac{-2}{\sqrt{1-0^2}+\sqrt{1+0^2}}$
$=\frac{-2}{1+1}$
$=-1$
View full question & answer→Question 192 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x}\right]$
Answer$ \lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x}\right]$
$=\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x} \times \frac{\sqrt{6+x+x^2}+\sqrt{6}}{\sqrt{6+x+x^2}+\sqrt{6}}\right] $
...[By rationalization]
$ =\lim _{x \rightarrow 0} \frac{\left(6+x+x^2\right)-6}{x\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\lim _{x \rightarrow 0} \frac{x+x^2}{x\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\lim _{x \rightarrow 0} \frac{x(1+x)}{x\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\lim _{x \rightarrow 0} \frac{1+x}{\sqrt{6+x+x^2}+\sqrt{6}} \quad \ldots[\because x \rightarrow 0, x \neq 0]$
$=\frac{\lim _{x \rightarrow 0}(1+x)}{\lim _{x \rightarrow 0}\left(\sqrt{6+x+x^2}+\sqrt{6}\right)}$
$=\frac{(1+0)}{\sqrt{6}+\sqrt{6}}$
$=\frac{1}{2 \sqrt{6}} $
View full question & answer→Question 202 Marks
Evaluate the following limits:
$\lim _{\Delta x \rightarrow 0}\left[\frac{(x+\Delta x)^2-2(x+\Delta x)+1-\left(x^2-2 x+1\right)}{\Delta x}\right]$
Answer$ \lim _{\Delta x \rightarrow 0}\left[\frac{(x+\Delta x)^2-2(x+\Delta x)+1-\left(x^2-2 x+1\right)}{\Delta x}\right]$
$=\lim _{\Delta x \rightarrow 0} \frac{x^2+2 x \cdot \Delta x+(\Delta x)^2-2 x-2 \Delta x+1-x^2+2 x-1}{\Delta x}$
$=\lim _{\Delta x \rightarrow 0} \frac{2 x \cdot \Delta x+(\Delta x)^2-2 \Delta x}{\Delta x}$
$=\lim _{\Delta x \rightarrow 0} \frac{\Delta x(2 x+\Delta x-2)}{\Delta x}$
$=\lim _{\Delta x \rightarrow 0}(2 x+\Delta x-2) \quad \ldots[\because \Delta x \rightarrow 0, \Delta x \neq 0]$
$=2 x+0-2$
$=2 x-2 $
View full question & answer→Question 212 Marks
Evaluate the following limits:
$\lim _{u \rightarrow 1}\left[\frac{u^4-1}{u^3-1}\right]$
Answer$ \lim _{u \rightarrow 1}\left[\frac{u^4-1}{u^3-1}\right]$
$=\lim _{u \rightarrow 1} \frac{\frac{u^4-1^4}{u-1}}{\frac{u^3-1^3}{u-1}} \quad \cdots\left[ \because u \rightarrow 1, u \neq 1,]$
$\therefore u-1 \neq 0 \right]$
$=\frac{\lim _{u \rightarrow 1}\left(\frac{u^4-1^4}{u-1}\right)}{\lim _{u \rightarrow 1}\left(\frac{u^3-1^3}{u-1}\right)}$
$=\frac{4(1)^3}{3(1)^2} \quad \ldots\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=\text { na"-1 }\right]$
$=\frac{4}{3} $
View full question & answer→Question 222 Marks
Evaluate the following limits:
$\lim _{x \rightarrow-2}\left[\frac{-2 x-4}{x^3+2 x^2}\right]$
Answer$ \lim _{x \rightarrow-2}\left[\frac{-2 x-4}{x^3+2 x^2}\right]$
$=\lim _{x \rightarrow-2} \frac{-2(x+2)}{x^2(x+2)}$
$=\lim _{x \rightarrow-2} \frac{-2}{x^2} \quad \cdots\left[ \because x \rightarrow-2, x \neq-2,7$
$\therefore x+2 \neq 0 \right]$
$=\frac{\lim _{x \rightarrow-2}(-2)}{\lim _{x \rightarrow-2}\left(x^2\right)}$
$=\frac{(-2)}{(-2)^2}$
$=\frac{-2}{4}$
$=\frac{-1}{2} $
View full question & answer→Question 232 Marks
Evaluate the following limits:
$\lim _{y \rightarrow 0}\left[\frac{5 y^3+8 y^2}{3 y^4-16 y^2}\right]$
Answer$ \lim _{y \rightarrow 0}\left[\frac{5 y^3+8 y^2}{3 y^4-16 y^2}\right]$
$=\lim _{y \rightarrow 0} \frac{y^2(5 y+8)}{y^2\left(3 y^2-16\right)}$
$=\lim _{y \rightarrow 0} \frac{5 y+8}{3 y^2-16} \quad \ldots\left[ \because y \rightarrow 0, y \neq 0,$
$\therefore y^2 \neq 0 \right]$
$=\lim _{y \rightarrow 0} \frac{(5 y+8)}{\left(3 y^2-16\right)}$
$=\frac{5(0)+8}{3(0)^2-16}$
$=\frac{8}{-16}$
$=-\frac{1}{2} $
View full question & answer→Question 242 Marks
In the following examples, given $\in > 0,$ find a $\delta > 0$ such that whenever, $|x – a| < \delta,$ we must have $|f(x) – l| < \in.
\lim _{x \rightarrow-3}(3 x+2)=-7$
AnswerWe have to find some $\delta$ so that
$\lim _{x \rightarrow-3}(3 x+2)=-7$
Here $a=-3, I=-7$ and $f(x)=3 x+2$
Consider $\in>0$ and $|\mathrm{f}(\mathrm{x})-\||<\in$
$ \therefore|3 x+2-(-7)|<\in$
$\therefore|3 x+9|<\in$
$\therefore|3(x+3)|<\in$
$\therefore|x+3|<\frac{\in}{3}$
$\therefore \delta<\frac{\in}{3} \text { such that }$
$|x+3| \leq \delta \Rightarrow|f(x)+7|<\in $
View full question & answer→Question 252 Marks
In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.$\lim _{x \rightarrow 2}(2 x+3)=7$
AnswerWe have to find some $\delta$ so that
$\lim _{x \rightarrow 2}(2 x+3)=7$
Here $a=2,1=1$ and $f(x)=2 x+3$
Consider $\in>0$ and $|\mathrm{f}(\mathrm{x})-\||<\in$
$ \therefore|(2 x+3)-7|<\in$
$\therefore|2 x+4|<\in$
$\therefore 2(x-2) \mid<\in$
$\therefore|x-2|<\frac{\in}{2}$
$\therefore \delta \leq \frac{\in}{2} \text { such that }$
$|2 x+4|<\delta \Rightarrow|f(x)-7|<\in $
$|2 x+4|<\delta \Rightarrow|f(x)-7|<\in$
View full question & answer→Question 262 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 7}\left[\frac{x^3-343}{\sqrt{x}-\sqrt{7}}\right]$
Answer$ \lim _{x \rightarrow 7} \frac{x^3-343}{\sqrt{x}-\sqrt{7}}$
$=\lim _{x \rightarrow 7} \frac{x^3-7^3}{x^{\frac{1}{2}}-7^{\frac{1}{2}}}$
$=\frac{\lim _{x \rightarrow 7} \frac{x^3-7^3}{x-7}}{\lim _{x \rightarrow 7} \frac{x^{\frac{1}{2}}-7^{\frac{1}{2}}}{x-7}}$
$\cdots\left[ \because x \rightarrow 7, x \neq 7$
$\therefore x-7 \neq 0 \right]$
$=\frac{3(7)^2}{\frac{1}{2}(7)^{-\frac{1}{2}}}$
$=6(49) \times 7^{\frac{1}{2}}$
$=294 \sqrt{7}$
View full question & answer→Question 272 Marks
Evaluate the following limits:
$\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]$
Answer$ \lim _{z \rightarrow a} \frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}$
$=\lim _{x \rightarrow a} \frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{(z+2)-(a+2)} $ Put $\mathrm{z}+2=y$ and $\mathrm{a}+2=\mathrm{b}$
As $\mathrm{z} \rightarrow \mathrm{a}, \mathrm{z}+2 \rightarrow \mathrm{a}+2$, i.e., $y \rightarrow \mathrm{b}$
$ \therefore \quad \lim _{x \rightarrow 2} \frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}$
$=\lim _{y \rightarrow b} \frac{y^{\frac{3}{2}}-b^{\frac{3}{2}}}{y-b}$
$=\frac{3}{2} \cdot \mathrm{b}^{\frac{1}{2}} \quad \ldots\left[\because \lim _{x \rightarrow a} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{n-1}\right]$
$=\frac{3}{2} \sqrt{a+2} \quad \ldots[\because \mathrm{b}=\mathrm{a}+2]$
View full question & answer→Question 282 Marks
Evaluate the following limits: $\lim _{x \rightarrow 0}\left[\frac{(1-x)^8-1}{(1-x)^2-1}\right]$
AnswerPut $1-x=y$
As $x \rightarrow 0, y \rightarrow 1$
$\therefore \lim _{x \rightarrow 0} \frac{(1-x)^8-1}{(1-x)^2-1}$
$ =\lim _{y \rightarrow 1} \frac{y^8-1^8}{y^2-1^2}$
$=\lim _{y \rightarrow 1} \frac{\frac{y^8-1^8}{y-1}}{\frac{y^2-1^2}{y-1}} \\ \cdots\left[\begin{array}{l} \because y \rightarrow 1, y \neq 1 \\ \therefore y-1 \neq 0 \end{array}\right]$
$=\frac{\lim _{y \rightarrow 1} \frac{y^8-1^8}{y-1}}{\lim _{y \rightarrow 1} \frac{y^2-1^2}{y-1}}$
$=\frac{8(1)^7}{2(1)^1} \quad \cdots\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n . a^{n-1}\right]$
$=4 $
View full question & answer→Question 292 Marks
Evaluate the following limits:
If $\lim _{x \rightarrow 5}\left[\frac{x^k-5^k}{x-5}\right]=500$, find all possible values of $k$.
Answer$ \lim _{x \rightarrow 5} \frac{x^k-5^k}{x-5}=500$
$\therefore \quad \mathrm{k}(5)^{\mathrm{k}-1}=500 \quad \cdots\left[\because \lim _{x \rightarrow \mathrm{a}} \frac{x^{\mathrm{n}}-\mathrm{a}^{\mathrm{a}}}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{\mathrm{n}-1}\right]$
$\therefore \quad \mathrm{k}(5)^{\mathrm{k}-1}=4 \times 125$
$\therefore \quad \mathrm{k}(5)^{\mathrm{k}-1}=4 \times(5)^3$
$\therefore \quad k(5)^{k-1}=4 \times(5)^{4-1}$
Comparing both sides, we get
$k=4$
View full question & answer→Question 302 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]$
Answer$ \lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]$
$=\lim _{x \rightarrow 7}\left[\frac{\left(x^{\frac{1}{3}}-7^{\frac{1}{3}}\right)\left(x^{\frac{1}{3}}+7^{\frac{1}{3}}\right)}{x-7}\right]$
$=\lim _{x \rightarrow 7}\left[\frac{x^{\frac{2}{3}}-7^{\frac{2}{3}}}{x-7}\right] \quad \ldots\left[\because(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})=\mathrm{a}^2-\mathrm{b}^2\right]$
$=\frac{2}{3}(7)^{\frac{-1}{3}} \quad \ldots\left[\because \lim _{x \rightarrow \infty} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{n-1}\right]$
$=\frac{2}{3} \cdot \frac{1}{7^{\frac{1}{3}}}$
$=\frac{2}{3 \sqrt[3]{7}} $
View full question & answer→Question 312 Marks
Evaluate the following limits:
$\lim _{x \rightarrow 1}\left[\frac{x+x^2+x^3+\ldots \ldots \ldots+x^n-n}{x-1}\right]$
Answer$ \lim _{x \rightarrow 1}\left[\frac{x+x^2+x^3+\ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{x+x^2+x^3+\ldots+x^{\mathrm{n}}-(1+1+1+\ldots \mathrm{n} \text { times })}{x-1}\right]$
$=\lim _{x \rightarrow 1} \frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right)+\ldots+\left(x^{\mathrm{n}}-1\right)}{x-1}$
$=\lim _{x \rightarrow 1}\left[\frac{x^1-1^1}{x-1}+\frac{x^2-1^2}{x-1}+\frac{x^3-1^3}{x-1}+\ldots+\frac{x^{\mathrm{n}}-1^{\mathrm{n}}}{x-1}\right]$
$=1(1)^0+2(1)^1+3(1)^2+4(1)^3+\ldots+\mathrm{n}(1)^{\mathrm{n}-1}$
$\ldots\left[\because \because \lim _{x \rightarrow 1} \frac{x^n-\mathrm{a}^{\mathrm{n}}}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{\mathrm{n}-1}\right]$
$=1+2+3+4+\ldots+\mathrm{n}$
$=\frac{\mathrm{n}(\mathrm{n}+1)}{2} $
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