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  • Using the method of L' Hospital, evaluate the limit. <img alt="\lim _{x \rightarrow 0} \frac{\left(\sqrt{e^{x}}\right)^{5}-1}{\sqrt{8^{x}}-1}" src="https://entrancecorner.oncodecogs.com/gi

Using the method of L' Hospital, evaluate the limit. \lim _{x \rightarrow 0} \frac{\left(\sqrt{e^{x}}\right)^{5}-1}{\sqrt{8^{x}}-1}

Option: 1

\frac{5}{6 \ln 2}


Option: 2

\frac{5}{3 \ln 2}


Option: 3

\frac{1}{3 \ln 2}


Option: 4

\frac{1}{\ln 2}


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(\sqrt{e^{x}}\right)^{5}-1}{\sqrt{8^{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(e^{a x}\right)=a e^{a x} \\ & \frac{d}{d x}\left(b^{x}\right)=b^{x} \ln b \end{aligned}

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(\sqrt{e^{x}}\right)^{5}-1}{\sqrt{8^{x}}-1} \\ & =\lim _{x \rightarrow 0} \frac{e^{\frac{5 x}{2}}-1}{8^{\frac{x}{2}}-1} \\ & =\lim _{x \rightarrow 0} \frac{\frac{d x}{d}\left(e^{\frac{5 x}{2}}-1\right)}{d x}\left(8^{\frac{x}{2}}-1\right) \\ \end{aligned}\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{5}{2} e^{\frac{5 x}{2}}}{\frac{8^{\frac{x}{2}} \ln 8}{2}}\left[ \frac{d}{d x}\left(e^{a x}\right)=a e^{a x}, \frac{d}{d x}\left(b^{x}\right)=b^{x} \ln b\right] \\ & =\frac{5 \times 1}{1 \times 3 \ln 2} \\ & =\frac{5}{3 \log 2} \quad\left[ \ln a=\log _{e} a\right] \end{aligned}

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Ritika Jonwal

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