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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow \infty} \frac{\log _{e} f(x)}{x}$, where $\lim _{x \rightarrow 0} f(x)=0" src="https://entrancecorn

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

\lim _{x \rightarrow \infty} \frac{\log _{e} f(x)}{x}$, 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 } \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 \infty} \frac{\log _{e} f(x)}{x}=\text { indeterminate }\left[\otimes \lim _{x \rightarrow \infty} f(x)=\infty\right]

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

\frac{d}{d x}\left(\log _{e} x\right)=\frac{1}{x}
 

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\log _{e} f(x)}{x} \\ & =\lim _{x \rightarrow \infty} \frac{\frac{d}{d x}\left(\log _{e} f(x)\right)}{\frac{d}{d x}(x)} \\ & =\lim _{x \rightarrow \infty} \frac{\frac{1}{f(x)}}{1} \quad\left[\Downarrow \frac{d}{d x}\left(\log _{e} x\right)=\frac{1}{x}\right] \\ & =\frac{0}{1}\left[\emptyset \lim _{x \rightarrow \infty} f(x)=\infty\right] \\ & =0 \end{aligned}

 

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manish

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