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  • Evaluate the following: <img alt="\mathrm{ \lim _{x \rightarrow \infty} x\left(\left(\frac{x}{x+1}\right)^x-\frac{1}{e}\right) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

Evaluate the following: \mathrm{ \lim _{x \rightarrow \infty} x\left(\left(\frac{x}{x+1}\right)^x-\frac{1}{e}\right) }

Option: 1

\mathrm{1 / 2 e}


Option: 2

\mathrm{2 e}


Option: 3

0


Option: 4

1


Answers (1)

best_answer

\mathrm{\begin{aligned} \left(\frac{x}{x+1}\right)^x & =\left(1+\frac{1}{x}\right)^{-x} \\ & =(1+t)^{-1 / t} \quad\left(t=\frac{1}{x}\right) \\ & =\exp \left(-\frac{\ln (1+t)}{t}\right) \\ & =\exp \left(-\frac{t-t^2 / 2+o\left(t^2\right)}{t}\right) \quad(\text { as } t \rightarrow 0) \\ & =\exp \left(-\left(1-\frac{t}{2}+o(t)\right)\right) \\ & =\exp \left(-1+\frac{t}{2}+o(t)\right) \\ & =\frac{1}{e} \exp \left(\frac{t}{2}+o(t)\right) \\ & =\frac{1}{e}\left(1+\frac{t}{2}+o(t)\right) \quad(\text { as } t \rightarrow 0) \\ & =\frac{1}{e}+\frac{t}{2 e}+o(t) \end{aligned}}

Hence our limit becomes 

\mathrm{\begin{aligned} \lim _{x \rightarrow \infty} x\left(\left(\frac{x}{x+1}\right)^x-\frac{1}{e}\right) & =\lim _{t \rightarrow 0^{+}} \frac{1}{t}\left(\frac{1}{e}+\frac{t}{2 e}+o(t)-\frac{1}{e}\right) \\ & =\lim _{t \rightarrow 0^{+}} \frac{1}{t}\left(\frac{t}{2 e}+o(t)\right) \\ & =\lim _{t \rightarrow 0^{+}}\left(\frac{1}{2 e}+o(1)\right) \\ & =\frac{1}{2 e} \end{aligned}}

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HARSH KANKARIA

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