Calculate this limit as x goes to infinity What is the following limit?
$$
\lim_{x \to \infty}
{\displaystyle{\large\left(1 + x\right)\left(1 + 1/x\right)^{2x} - x\,{\rm e}^{2}}
\over \displaystyle{\large{\rm e}^{2} - \left(1 + 1/x\right)^{2x}}}
$$
Thanks.

Define the numerator and the denominator of the limit as:
$$a(x):={\large\left(1 + x\right)\left(1 + 1/x\right)^{2x} - x\,{\rm e}^{2}}$$
$$b(x):={\large{\rm e}^{2} - \left(1 + 1/x\right)^{2x}}$$
Then
$$A(t)=a(1/t)={\large\left(1 + (1/t)\right)\left(1 + t\right)^{2/t} - (1/t)\,{\rm e}^{2}}$$
$$B(t)=b(1/t)={\large{\rm e}^{2} - \left(1 + t\right)^{2/t}}$$
We can now expand $A(t)$ and $B(t)$ at $t=0$ ($x=\infty$):
$$A(t)=\frac{{\rm e}^{2}}{6}t+O(t^2)$$
$$B(t)={\rm e}^{2}t+O(t^2)$$
Thus
$$\lim_{x \to \infty}\frac{a(x)}{b(x)}=\lim_{t \to 0}\frac{A(t)}{B(t)}
=\lim_{t \to 0}\frac{\frac{{\rm e}^{2}}{6}t}{{\rm e}^{2}t}=\frac{1}{6}$$
It is exactly as what Mark Fischler predicted.
