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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\log (1+x)^{\sin a}}{\sqrt{1+x}-1}, a \neq 0" src="https://entrancec

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

\lim _{x \rightarrow 0} \frac{\log (1+x)^{\sin a}}{\sqrt{1+x}-1}, a \neq 0

Option: 1

\frac{\cos 2 a}{\sin a}


Option: 2

\frac{1-\cos 2 a}{\sin a}


Option: 3

\frac{1-\cos a}{\sin2 a}


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}, \text { 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

\lim _{x \rightarrow 0} \frac{\log (1+x)^{\sin a}}{\sqrt{1+x}-1} =\frac{0}{0}

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

\begin{aligned} & \frac{d}{d x}\left(\log _e x\right)=\frac{1}{x} \\ & \frac{d}{d x} x^n=n x^{n-1} \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\log (1+x)^{\sin a}}{\sqrt{1+x}-1} \\ & =\lim _{x \rightarrow 0} \frac{\sin a \log (1+x)}{\sqrt{1+x}-1} \\ & =(\sin a) \lim _{x \rightarrow 0} \frac{\frac{d}{d x}(\log (1+x))}{\frac{d x}{d x}(\sqrt{1+x}-1)} \\ & \end{aligned}

\begin{aligned} & =(\sin a) \lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{\frac{1}{2 \sqrt{1+x}}} \quad\left[\mathrm{~N} \frac{d}{d x} x^n=n x^n+\frac{d}{d x}\left(\log _e x\right)=\frac{1}{x}\right] \\ & =(\sin a) \frac{1}{\frac{1+0}{2 \sqrt{1+0}}} \\ & =2 \sin a \\ & =\frac{2 \sin ^2 a}{\sin a} \\ & =\frac{1-\cos 2 a}{\sin a} \end{aligned}

 

Posted by

Shailly goel

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