# Find the limit $\displaystyle\lim_{n\rightarrow\infty}{(1+1/n)^{n^2}e^{-n}}$?

Find the limit $\displaystyle\lim_{n\rightarrow\infty}{(1+1/n)^{n^2}e^{-n}}$? I found the limit as $e^{-1/2}$ using l'Hospital rule. I guess I made a mistake. Because the limit seems to be 1. Also, can we find the limit without L'Hospital rule?

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The limit only "seems to be $1$" if you naively substitute $(1+\frac1n)^n\approx e$, hence $(1+\frac1n)^{n^2}\approx e^n$. But the $\approx$ is not good enough to overcome the growing exponent ... –  Hagen von Eitzen Jun 17 '13 at 16:21
When you found $e^{-1/2}$, did you ln both sides? $\ln(e^{-1/2})=-\frac{1}{2}$, so you were really close to getting the answer. –  Sujaan Kunalan Jun 17 '13 at 16:30
@HagenvonEitzen, i guess it would be true if $x^n$ were uniform convergence on the interval $[1,3]$. –  mtm Jun 17 '13 at 17:32

Let $$y=\left(1+\frac1n\right)^{n^2}\cdot e^{-n}$$

$$\implies \ln y=n^2\ln\left(1+\frac1n\right)-n$$

$$\text{Putting }n=\frac1h, \lim_{n\to\infty}\ln y=\lim_{h\to0}\frac{\ln(1+h)-h}{h^2}\text{ which is of the form } \frac00$$

Applying L'Hospitals Rule, $$\lim_{h\to0}\frac{\ln(1+h)-h}{h^2}=\lim_{h\to0}\frac{\frac1{1+h}-1}{2h}=\lim_{h\to0}\frac{1-(1+h)}{2h(1+h)}=-\frac12$$ as $h\ne0$ as $h\to0$

Alternatively, using Taylor expansion of $\ln(1+x)$

$$\lim_{h\to0}\frac{\ln(1+h)-h}{h^2}=\lim_{h\to0}\frac{h-\frac{h^2}2+O(h^3)-h}{h^2}=-\frac12$$

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@mtm, so you are right. Btw, would you please your method –  lab bhattacharjee Jun 17 '13 at 16:28
i applied two times L' Hospital. –  mtm Jun 17 '13 at 16:33
@mtm, its $\lim_{n\to\infty}\ln y=-\frac12\iff \lim_{n\to\infty}y=e^{-\frac12}$ –  lab bhattacharjee Jun 17 '13 at 16:38