Solve the following Question.(1 Marks)

Question 11 Mark

Evaluate the following limits:
$\lim _{\theta \rightarrow 0}\left[\frac{1-\cos 2 \theta}{\theta^2}\right]$

Answer

$ \lim _{\theta \rightarrow 0} \frac{1-\cos 2 \theta}{\theta^2}=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 \theta}{\theta^2}$
$=2 \lim _{\theta \rightarrow 0}\left(\frac{\sin \theta}{\theta}\right)^2=2(1)^2=2$

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Question 21 Mark

Evaluate the following limits:
$\lim _{x \rightarrow 2}\left[\frac{x^3-4 x^2+4 x}{x^2-1}\right]$

Answer

$ \lim _{x \rightarrow 2}\left[\frac{x^3-4 x^2+4 x}{x^2-1}\right]$
$=\frac{\lim _{x \rightarrow 2}\left(x^3-4 x^2+4 x\right)}{\lim _{x \rightarrow 2}\left(x^2-1\right)}$
$=\frac{2^3-4(2)^2+4(2)}{2^2-1}$
$=\frac{0}{3}=0 $

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Question 31 Mark

Evaluate the following limits : $\lim _{x \rightarrow 3}\left[\frac{x^2+2 x-15}{x^2-5 x+6}\right]$

Answer

$\lim _{x \rightarrow 3}\left[\frac{x^2+2 x-15}{x^2-5 x+6}\right]$
$=\lim _{x \rightarrow 3} \frac{(x+5)(x-3)}{(x-2)(x-3)}$
$=\lim _{x \rightarrow 3} \frac{x+5}{x-2} \quad \cdots\left[\begin{array}{l} \because x \rightarrow 3, x \neq 3, \\ \therefore x-3 \neq 0
\end{array}\right]$
$=\frac{\lim _{x \rightarrow 3}(x+5)}{\lim _{x \rightarrow 3}(x-2)}=\frac{3+5}{3-2}$
$=8$

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Question 41 Mark

Evaluate the following limits : $\lim _{x \rightarrow-3}\left[\frac{x+3}{x^2+4 x+3}\right]$

Answer

$\lim _{x \rightarrow-3}\left[\frac{x+3}{x^2+4 x+3}\right]$
$=\lim _{x \rightarrow-3} \frac{x+3}{(x+3)(x+1)} \quad \quad \cdots\left[\begin{array}{l} \because x \rightarrow-3, x \neq-3, \\ \therefore x+3 \neq 0 \end{array}\right]$
$=\lim _{x \rightarrow-3} \frac{1}{x+1}$
$=\frac{1}{-3+1}$
$=-\frac{1}{2}$

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Question 51 Mark

Evaluate the following limits : $\lim _{z \rightarrow 2}\left[\frac{z^2-5 z+6}{z^2-4}\right]$

Answer

$\lim _{x \rightarrow 2} \frac{z^2-5 z+6}{z^2-4}$
$=\lim _{x \rightarrow 2} \frac{(z-3)(z-2)}{(z+2)(z-2)}$
$=\lim _{x \rightarrow 2} \frac{z-3}{z+2} \quad \ldots\left[\begin{array}{l} \because z \rightarrow 2, z \neq 2, \\ \therefore z-2 \neq 0 \end{array}\right]$
$=\frac{\lim _{z \rightarrow 2}(z-3)}{\lim _{x \rightarrow 2}(z+2)}$
$=\frac{2-3}{2+2}$
$=-\frac{1}{4}$

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Question 61 Mark

Evaluate the following limits : $\lim _{x \rightarrow 5}\left[\frac{x^3-125}{x^5-3125}\right]$

Answer

$\lim _{x \rightarrow 5} \frac{x^3-125}{x^5-3125}$
$=\lim _{x \rightarrow 5} \frac{\left(\frac{x^3-5^3}{x-5}\right)}{\left(\frac{x^3-5^5}{x-5}\right)} \quad \cdots\left[\begin{array}{l} \because x \rightarrow 5, x \neq 5, \\ \therefore x-5 \neq 0 \end{array}\right]$
$=\frac{\lim _{x \rightarrow 5} \frac{x^3-5^3}{x-5}}{\lim _{x \rightarrow 5} \frac{x^5-5^5}{x-5}}$
$=\frac{3(5)^2}{5(5)^4} \quad \cdots\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n \cdot a^{n-1}\right]$
$=\frac{3}{(5)^3}$
$=\frac{3}{125}$

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Question 71 Mark

Evaluate the following limits:

$
\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]
$

Answer

$
\begin{aligned}
\lim _{x \rightarrow 2} \frac{x^{-3}-2^{-3}}{x-2} & =(-3) \cdot(2)^{-4} \\
& \cdots\left[\because \lim _{x \rightarrow a} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{n} \cdot \mathrm{a}^{n-1}\right] \\
& =-3 \times \frac{1}{2^4} \\
& =\frac{-3}{16}
\end{aligned}
$

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Question 81 Mark

Evaluate the following limits:

$\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+6}}{x}\right]$

Answer

$
\begin{aligned}
\lim _{x \rightarrow 3} \frac{\sqrt{2 x+6}}{x} & =\frac{\lim _{x \rightarrow 3} \sqrt{2 x+6}}{\lim _{x \rightarrow 3} x} \\
& =\frac{\sqrt{2(3)+6}}{3} \\
& =\frac{\sqrt{12}}{3}=\frac{2 \sqrt{3}}{3}
\end{aligned}
$

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Question 91 Mark

Evaluate the following limits: $\lim _{z \rightarrow-5}\left[\frac{\left(\frac{1}{z}+\frac{1}{5}\right)}{z+5}\right]$

Answer

$\lim _{x \rightarrow-5} \frac{\frac{1}{z}+\frac{1}{5}}{z+5}$
$=\lim _{x \rightarrow-5} \frac{\frac{5+z}{z+5}}{z+5}$
$=\lim _{x \rightarrow-5}\left[\frac{z+5}{5 z(z+5)}\right]$
$=\lim _{x \rightarrow-5} \frac{1}{5 z} \quad \cdots\left[\begin{array}{l} \because z \rightarrow-5, \quad z \neq-5, \\
\therefore z+5 \neq 0 \end{array}\right]$
$=\frac{1}{5(-5)}$
$=\frac{-1}{25}$

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Question 101 Mark

Evaluate the following limits: $\lim _{y \rightarrow-3}\left[\frac{y^5+243}{y^3+27}\right]$

Answer

$\lim _{y \rightarrow-3} \frac{y^5+243}{y^3+27}$
$=\lim _{y \rightarrow-3} \frac{\left(\frac{y^5+243}{y+3}\right)}{\left(\frac{y^3+27}{y+3}\right)} \quad \cdots\left[\begin{array}{l} \because y \rightarrow-3, y \neq-3, \\ \therefore y+3 \neq 0 \end{array}\right]$
$=\frac{\lim _{y \rightarrow-3}\left[\frac{y^5-(-3)^5}{y-(-3)}\right]}{\lim _{y \rightarrow-3}\left[\frac{y^3-(-3)^3}{y-(-3)}\right]} \quad \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{x^n- a ^n}{x- a \cdot}= n . a ^{n-1}\right]$
$=\frac{5(-3)^4}{3(-3)^2}$
$=\frac{5}{3} \times 9$
$=15$

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Question 111 Mark

Evaluate the following limits: $\lim _{z \rightarrow-3}\left[\frac{\sqrt{Z+6}}{Z}\right]$

Answer

$\lim _{x \rightarrow-3}\left[\frac{\sqrt{z+6}}{z}\right] =\frac{\lim _{z \rightarrow-3} \sqrt{z+6}}{\lim _{x \rightarrow-3} z}$
$ =\frac{\sqrt{-3+6}}{-3}$
$=\frac{\sqrt{3}}{-3}$
$=\frac{-1}{\sqrt{3}}$

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Maths Solve the following Question.(1 Marks)

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