Calculus Solutions Ex# 3.6

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Q 1-2. Evaluaté the given limit without using L’Hopital rule, and then check that your answer is correct using L’Hopital rule. :
1. (a)

    \[\lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{2}+2 x-8}\]

    \[\begin{aligned} & \lim_{x \rightarrow 2} \frac{(x-2)(x+2)}{x+4 x-2 x-8}=\lim_{x \rightarrow 2} \frac{(x-2)(x+2)}{x(x+4)-2(x+4)} \\ \\ \Rightarrow \quad & \lim_{x \rightarrow 2} \frac{(x-2)(x+2)}{(x+4)(x-2)}=\lim_{x \rightarrow 2} \frac{x+2}{x+1} \cdot \frac{2+2}{2+3}=\frac{4}{6}=\frac{2}{3} \end{aligned}\]

and Solution using L’Hopital rule

    \[\lim_{x \rightarrow 2} \frac{x^{2}-4}{x^{2}+2 x-8}=\frac{2^{2}-4}{2^{2}+2(2)-8}=\frac{4-4}{4+4-8}=\frac {0}{0} \text \quad {form}\]

Applying L’Hopital rule

    \[\begin{aligned} & \lim_{x \rightarrow 2} \frac{x^{2}-4}{x^{2}+2 x-8}= \lim_{x \rightarrow 2} \frac{\frac{d}{d x}\left[x^{2}-4\right]}{\frac{d}{d x}\left[x^{2}+2 x-8\right]} \\ \\ & =\lim_{x \rightarrow 2} \frac{2 x}{2 x+2}=\frac{2(2)}{2(2)+2}=\frac{4}{6}=\frac{2}{3} \\ \end{aligned}\]

1 (b)

    \[\begin{aligned} & \lim_{x \rightarrow+\infty} \frac{2 x-5}{3 x+7}=\lim_{x \rightarrow+\infty} \frac{2 x}{3 x}=\lim_{x \rightarrow+\infty} \frac{2}{3}=\frac{2}{3} \end{aligned}\]

Solution using L ‘Hopital rule

    \[\lim_{x \rightarrow+\infty} \frac{2 x-5}{3 x+7}=\frac{\lim_{x \rightarrow+\infty}(2 x-5)}{\lim_{x \rightarrow+\infty}(3 x+7)}=\frac{2(\infty)-5}{3(\infty)+7}=\infty / \infty\]

Applying L ‘Hopital rule

    \[\lim _{x \rightarrow+\infty} \frac{2 x-5}{3 x+7}=\operatorname{Lim}_{x \rightarrow+\infty} \frac{\frac{d}{d x}[2 x-5]}{\frac{d}{d x}[3 x+7]}=\lim _{x \rightarrow+\infty} \frac{2}{3}=\frac{2}{3}\]

2. (a)

    \[\lim _{x \rightarrow 0} \frac{\sin x}{\tan x} }\]

    \[\lim_{x \rightarrow 0} \frac{\sin x}{sin x/cos x}=\lim_{x \rightarrow 0} \cos x=\cos (0)=1\]

Solution using L’Hopital rule.

    \[\lim_{x \rightarrow 0} \frac{\sin x}{\tan x}=\frac{\lim_{x \rightarrow 0} \sin x}{\lim_{x \rightarrow 0} \tan x}=\frac{\sin 0}{\tan0}=\frac{0}{0}\]

Applying L ‘Hopital rule. Differentiating numerator and denominator with respect to x.

    \[\lim_{x \rightarrow 0} \frac{\sin x}{\tan x}=\lim_{x \rightarrow 0} \frac{\frac{d}{d x}[\sin x]}{\frac{d}{d}[\tan x]}=\lim_{x \rightarrow 0} \frac{\cos x}{\sec ^{2} x}=\frac{\cos 0}{\sec ^{2}(0)}=\frac{1}{1}=1\]

(b)

    \[\begin{aligned} & \lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1} \\ \\ & \lim_{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1}=\lim_{x \rightarrow 1} \frac{(x-1)(x+1)}{(x-1)\left(x^{2}+x+1\right)} \\ \\ & \lim _{x \rightarrow 1} \frac{x+1}{x^{2}+x+1}=\frac{1+1}{1^{2}+1+1}=\frac{2}{3} \end{aligned}\]

Solution using L’Hopital rule.

    \[\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1}=\frac{1^{2}-1}{1^{3}-1}=\frac{0}{0}\]

Applying L’Hopital rule

    \[\lim_{x \rightarrow 1} \frac{x^{2}-1}{x^{3}-1}-\lim _{x \rightarrow 1} \frac{\frac{d}{d x}\left[x^{2}-1\right]}{\frac{d}{d x}\left[x^{3}-1\right]}=\lim _{x \rightarrow 1} \frac{2 x}{3 x^{2}}=\frac{2(1)}{3(1)^{2}}=\frac{2}{3}\]

Q.7-10 Find the limits
Solution:
(7)

    \[\lim _{x \rightarrow 0} \frac{e^{x}-1}{\sin x}=\frac{e^{0}-1}{\sin 0}=\frac{1-1}{0}=\frac {0}{0}\]

Applynig L’Hopital rule

    \[\begin{aligned} & \lim_{x \rightarrow 0} \frac{e^{x}-1}{\sin x}=\lim_{x \rightarrow 0} \frac{\frac{d}{d x}\left[e^{x}-1\right]}{\frac{d}{d x}[\sin x]}=\lim_{x \rightarrow 0} \frac{e^{x}}{\cos x} \\ \\ & =\frac{e^{0}}{\cos 0}=\frac{1}{1}=1 \end{aligned}\]

(8)

    \[\begin{enumerate} \setcounter{enumi}{7} \item $\lim_{x \rightarrow 0} \frac{\sin 2 x}{\sin 5 x}=\frac{\sin 0}{\sin 0}=\frac{0}{0}$ \end{enumerate}\]

Applying L’Hopital rule . Taking derivative of numerator and denominator with respect to x.

    \[\begin{aligned} \lim_{x \rightarrow 0} \frac{\sin 2 x}{\sin 5 x} & =\lim_{x \rightarrow 0} \frac{\frac{d}{d x}[\sin 2 x]}{\frac{d}{d x}[\sin 5 x]}=\lim_{x \rightarrow 0} \frac{2 \cos 2 x}{5 \cos 5 x} \\ \\ &=\frac{2 \cos 0}{5 \cos 0}\\ \\ & =\frac{2(1)}{5(1)} =2 / 5 \end{aligned}\]

(9)

    \[\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}= \frac{\tan 0}{0}=\frac{0}{0}\]

Applying L’Hopital rule. Taking derivative of numerator and denominator with respect to \theta

    \[\begin{aligned} & \lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=\lim_{\theta \rightarrow 0} \frac{\frac{d}{d \theta}[\tan \theta]}{\frac{d}{d \theta}(\theta)}\\ & =\lim_{\theta \rightarrow 0} \frac{\sec ^{2} \theta}{1}=\sec ^{2}(0)=1 \end{aligned}\]

(10)

    \[\lim_{t \rightarrow 0} \frac{t e^{t}}{1-t}=\frac{0 e^{0}}{1-e^{0}}=\frac{0}{1-1}=\frac{0}{0}\]

Applying L’Hopital rule. Differentiating both numerator and denominator with respect to t.

    \[\begin{aligned} \lim_{t \rightarrow 0} \frac{t e^{t}}{1-e^{t}}=\lim_{t \rightarrow 0} \frac{\frac{d}{d t}\left[t e^{t}\right]}{\frac{d}{d t}\left[1-e^{t}\right]} \\ \\ =\lim_{t \rightarrow 0} \frac{t \frac{d}{d t}\left[e^{t}\right]+e^{t} \frac{d}{d t}(t)}{0-e^{t}} \\ \\ =\lim_{t \rightarrow 0} \frac{t e^{t}+e^{t}(1)}{-e^{t}} \\ \\ =\lim_{t \rightarrow 0} \frac{t e^{t}+e^{t}}{-e^{t}} \\ \\ =\frac{0 e^{0}+e^{0}}{-e^{0}}=\frac{0+1}{-1}=-1 \end{aligned}\]