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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\left(e^{x}\right)^{5}-\left(\sqrt{e^{x}}\right)^{5}}{x}" src="https://entrancecorner.oncodecog

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\left(e^{x}\right)^{5}-\left(\sqrt{e^{x}}\right)^{5}}{x}

Option: 1

\pm 2.5


Option: 2

-2.5


Option: 3

5


Option: 4

2.5


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

\lim _{x \rightarrow 0} \frac{\left(e^{x}\right)^{5}-\left(\sqrt{e^{x}}\right)^{5}}{x}=\frac{0}{0}

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

\frac{d}{d x}\left(e^{a x}\right)=a e^{a x}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(e^{x}\right)^{5}-\left(\sqrt{e^{x}}\right)^{5}}{x} \\ & =\lim _{x \rightarrow 0} \frac{e^{5 x}-e^{\frac{5 x}{2}}}{x} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d}{d x}\left(e^{5 x}-e^{\frac{5 x}{2}}\right)}{\frac{d}{d x}(x)} \\ & =\lim _{x \rightarrow 0} \frac{5 e^{5 x}-\frac{5}{2} e^{\frac{5 x}{2}}}{1} \quad\left[ \frac{d}{d x}\left(e^{\alpha x}\right)=a e^{\alpha x}\right] \\ \end{aligned}\begin{aligned} & =\frac{5 \times 1-\frac{5}{2} \times 1}{1} \\ & =\frac{5}{2} \\ & =2.5 \end{aligned}

Posted by

Rakesh

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