Evaluating the limit $\lim_{x\to\infty}\frac{(1+1/x)^{x^2}}{e^x}$

Question

Could anybody show me step by step why the following equality holds? $$\lim_{x\to\infty}\frac{(1+1/x)^{x^2}}{e^x} = e^{-1/2}$$

The most obvious method gives you 1 as an answer, but I understand that only the limit of $(1+1/x)^x$ is $e$, but expression is not actually equal to $e$. And now I am stuck.

Answer 1

Compute the limit of the logarithm of your function: $$ \lim_{x\to\infty}\log\frac{(1+1/x)^{x^2}}{e^x}= \lim_{x\to\infty}(x^2\log(1+1/x)-x) $$ Now set $x=1/t$: $$ \lim_{x\to\infty}(x^2\log(1+1/x)-x)= \lim_{t\to0^+}\frac{\log(1+t)-t}{t^2} $$

The limit becomes $\displaystyle\lim_{t\to0^+}\frac{t-t^2/2+o(t^2)-t}{t^2}=-\frac{1}{2}$. So your original limit is $e^{-1/2}$.

Answer 2

Say $$y = \frac{\left( 1 + \frac{1}{x}\right)^{x^2}}{e^x}$$ Then $\log(y) =x^2 \log\left( 1 + \frac{1}{x}\right)-x$. We change variables $t=\frac{1}{x}$. Then $\log(y)=\frac{1}{t^2} \log(1+t) - \frac{1}{t}$.

Near $t=0$, $$\log(1+t) = t - \frac{t^2}{2}+\frac{t^3}{3}- \frac{t^4}{4} +\ldots$$

So near $t=0$, $$\log(y) = \left(\frac{1}{t} - \frac{1}{2} + O(t)\right) - \frac{1}{t} = - \frac{1}{2} + O(t) $$

Going back in terms of $x$, for large $x$

$$\log(y) = - \frac{1}{2} + O\left(\frac{1}{x}\right) $$

Hence $$\lim_{x\rightarrow\infty} \log(y) = -\frac{1}{2}$$

exponentiating we have the result.