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  • Using the method of L’ Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\log (1+x)}{\sqrt{49+x}-7}" src="https://entrancecorner.oncodecogs.c

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

\lim _{x \rightarrow 0} \frac{\log (1+x)}{\sqrt{49+x}-7}

Option: 1

\text { Q14 }


Option: 2

\frac{+}{-}14


Option: 3

-14


Option: 4

14


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)}{\sqrt{49+x}-7}=\frac{0}{0}

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

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

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\log (1+x)}{\sqrt{49+x}-7} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x} \log (1+x)}{\frac{d}{d x}(\sqrt{49+x}-7)} \\ & =\lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{\frac{1}{2 \sqrt{49+x}}} \quad\left[\frac{d}{d x} x^n=n x^{n-1}, \frac{d}{d x}\left(\log _8 x\right)=\frac{1}{x}\right] \\ & =\frac{\frac{1}{1+0}}{\frac{1}{2 \sqrt{49+0}}} \\ & =14 \\ & \end{aligned}

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