Finding $\lim_{x \to \infty}\, {\mathrm{e}}^{x}{\left(1+\frac{1}{x}\right)}^{x-{x}^{2}}$ 
$$\lim_{x \to \infty}\, {\mathrm{e}}^{x}{\left(1+\frac{1}{x}\right)}^{x-{x}^{2}}$$

I don't know where to start. Can I use series expansion of $\mathrm{e}^{x}$ and $(1+\frac{1}{x})\,$? Or L'Hôpital's rule since it is $\frac{\infty}{\infty}$ form. I am looking for hint to get started.
 A: Hint:
If $A=\displaystyle\lim_{x \to \infty}\, {\mathrm{e}}^{x}{\cdot}{\left(1+\dfrac{1}{x}\right)}^{x-{x}^{2}}$
$\ln A=\lim_{x\to\infty}x+(x-x^2)\ln(1+1/x)$
$$=\lim_{h\to0^+}\dfrac{h+(h-1)\ln(1+h)}{h^2}$$
Now apply L'Hospital or Series Expansion
A: Consider $$y=e^{x}\,{\left(1+\dfrac{1}{x}\right)}^{x-{x}^{2}}$$ Take logarithms
$$\log(y)=x +(x-x^2)\log\left(1+\dfrac{1}{x}\right)$$ Now use Taylor expansion to get
$$\log(y)=x +(x-x^2)\left(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\frac{1}{x^4}\right) \right)=\frac{3}{2}-\frac{5}{6 x}+O\left(\frac{1}{x^2}\right)$$ Use Taylor again
$$y=e^{\log(y)}=e^{\frac 32}\left(1-\frac{5}{6 x} \right)+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.
For illustartion purposes, use $x=10$; the exact value is $4.14575$ while the approximation would give $4.10821$ (less than $1$% error).
A: Let use Taylor's expansion $t \to 0$


*

*$\log (1+t)=t-\frac12t^2+O(t^3)$


and then
$${\left(1+\dfrac{1}{x}\right)}^{x-{x}^{2}}=e^{(x-x^2)\log{\left(1+\dfrac{1}{x}\right)}}=e^{(x-x^2)\left(\frac1x-\frac1{2x^2}+O(1/x^3)\right)}=e^{\frac32-x+O(1/x)}$$
A: As an alternative we have
$$e^x{\left(1+\dfrac{1}{x}\right)}^{x-{x}^{2}}=e^{x+(x-x^2)\log{\left(1+\dfrac{1}{x}\right)}}\to e^{3/2}$$
indeed without l'Hopital and Taylor's expansion we have
$$\lim_{t\to0}\frac{t-\log(1+t)}{t^2}=\frac12 \iff\log(1+t)\sim t-\frac12 t^2$$
and therefore
$${x+(x-x^2)\log{\left(1+\dfrac{1}{x}\right)}}\sim x+(x-x^2)\left(\frac1x-\frac1{2x^2}\right)=x+1-\frac1{2x}-x+\frac12=\frac32-\frac1{2x}\to \frac32$$
