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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\sin f(x)}{x}, \text { where } \lim _{x \rightarrow 0} f(x)=0" src="

Using the method of L’ Hospital, evaluate the limit.

\lim _{x \rightarrow 0} \frac{\sin f(x)}{x}, \text { where } \lim _{x \rightarrow 0} f(x)=0

Option: 1

\infty


Option: 2

0


Option: 3

1


Option: 4

 Cannot be determined


Answers (1)

best_answer

Apply the L' Hospital's Rule, when \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, etc. [an intermediate form] in the following way

- Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} exists.

- Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin f(x)}{x} \\ & =\frac{0}{0} \quad\left[ \lim _{x \rightarrow 0} f(x)=0\right] \end{aligned}

 

Here use the L' Hospital rule and also refer to the following formula.

\frac{d}{d x}(\sin x)=\cos x

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin f(x)}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\sin f(x))}{\frac{d}{d x}(x)} \\ & =\lim _{x \rightarrow 0} \frac{\cos f(x)}{1} \quad\left[\unrhd \frac{d}{d x}(\sin x)=\cos x\right] \\ & =\frac{1}{1}\left[\square \lim _{x \rightarrow 0} f(x)=0\right] \\ & =1 \end{aligned}

Posted by

manish painkra

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