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  • Without using the method of L' Hospital, find the limit.  <img alt="\lim _{x \rightarrow \infty}\left(x\left(\sqrt[x]{e^{2}}\right)^{5}-x\right)" src="https://entrancecorner.oncodecog

Without using the method of L' Hospital, find the limit.  \lim _{x \rightarrow \infty}\left(x\left(\sqrt[x]{e^{2}}\right)^{5}-x\right)

Option: 1

\ln 10


Option: 2

10


Option: 3

e^{10}


Option: 4

\ln 10


Answers (1)

best_answer

Note the following exponential limit.

\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1

Assume  \frac{1}{x}=t

So, as  x \rightarrow \infty, t \rightarrow 0

Now, evaluate the following limit using the above formula and the assumption.

\begin{aligned} & \lim _{x \rightarrow \infty}\left(x\left(\sqrt[x]{e^{2}}\right)^{5}-x\right) \\ & =\lim _{x \rightarrow \infty}\left(x e^{\frac{10}{x}}-x\right) \\ & =\lim _{x \rightarrow \infty} x\left(e^{\frac{10}{x}}-1\right) \\ \end{aligned}\begin{aligned} & =\lim _{t \rightarrow 0}\left(\frac{e^{10 t}-1}{t}\right) \\ & =10 \times \lim _{t \rightarrow 0}\left(\frac{e^{10 t}-1}{10 t}\right) \\ & =10 \times 1 \\ & =10\end{aligned}

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