Difficulty in evaluating a limit: $\lim_{x \to \infty} \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$

Question

The limit I have to evaluate is this - $$\lim_{x \to \infty} \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$$

I first checked if L'Hopital's rule applies here. The limit of both numerator and denominator is $\infty$. But differentiating the denominator yield a even more complicated expression.

I am not getting how to approach this question using some other method. Thank you.

Answer 1

let $$\begin{align}y &= \frac{e^x}{\left(1 + \frac 1 x\right)^{x^2}},\\\ \ln y &= x - x^2 \ln \left(1 + \frac 1x\right) \\ &= x - x^2\left(\frac 1x - \frac1 {2x^2} + \frac 1 {3x^3 }+\cdots\right)\\ &=\frac 12 - \frac 1 {3x} + \frac 1 {4x^2}+\cdots \rightarrow \frac 12 \end{align}$$

therefore $$\lim_{x \to \infty}\frac {e^x} {\left(1 + \frac 1 x\right)^{x^2}}= \sqrt e.$$

Answer 2

Hint

Define $$A= \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$$ and take logarithms of both sides; so $$\log(A)=x -x^2\log(1+\frac 1 x)$$ Now, using the fact that, by Taylor, for small $y$, $\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$, replace $y$ by $\frac 1x$ to get $$\log(A)=x-x^2\Big(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+\cdots\Big)=\frac{1}{2}-\frac{1}{3 x}+\cdots$$