evaluate lim: $\lim _{x\to \infty }\left(2x\left(e-\left(1+\frac{1}{x}\right)^x\right)\right)$ I was trying to solve the above limit and it seems like I'm getting mixed results.
I used the fact that:
$\lim _{x\to \infty }\left(1+\frac{1}{x}\right)^x = e$
And after that trying with L'Hospital's rule but it didn't get me much further.
 A: HINT: rewrite your term in the form $$\frac{e-\left(1+\frac{1}{x}\right)^{x}}{\frac{1}{2x}}$$ and apply the rules of L'Hospital
A: With some work, l'Hopital's rule solves the problem.
The hard part is the derivative of $x\mapsto\left(1+\frac1x\right)^x$. For this, write
$$y=\left(1+\frac1x\right)^x$$
$$\ln y=x\ln\left(1+\frac1x\right)$$
Now take derivatives:
$$\frac{y'}y=\ln\left(1+\frac1x\right)-\frac1{x+1}$$
and hence
$$y'=\left(1+\frac1x\right)^x\left[\ln\left(1+\frac1x\right)-\frac1{x+1}\right]$$
Now, l'Hopital:
$$\lim_{x\to\infty}2x\left[e-\left(1+\frac1x\right)^x\right]=2\lim_{x\to\infty}x^2\left(1+\frac1x\right)^x\left[\ln\left(1+\frac1x\right)-\frac1{x+1}\right]$$
With more l'Hopital and some patience you can find out that
$$\lim_{x\to\infty}x^2\left[\ln\left(1+\frac1x\right)-\frac1{x+1}\right]=\frac12$$
Thus, the final result is $e$.
Warning: I have made all this by hand, so I advise to double check the result.
A: I'll make use of the following Taylor approximations below: For small $h>0,$ $$\ln (1+h) = h - h^2/2 + O(h^3), e^h = 1 + h +O(h^2).$$
Rewrite our expression as $2xe(1-(1+1/x)^x/e).$ Now
$$\ln[(1+1/x)^x/e] = x\ln (1+1/x) - 1 = x(1/x - 1/2x^2 + O(1/x^3)) - 1 = -1/2x +O(1/x^2).$$
Exponentiating back gives 
$$(1+1/x)^x/e = 1-1/2x + O(1/x^2).$$
Thus 
$$2xe(1-(1+1/x)^x/e) = 2xe(1/2x + O(1/x^2))=e + O(1/x).$$
So the limit is $e.$
A: Make the substitution $x \to 1/x$.  Thus, 
$$\begin{align}
\lim_{x\to \infty}\left(2x\left(e-\left(1+\frac1x\right)^x\right)\right)&=\lim_{x\to 0}2\frac{e-\left(1+x\right)^{1/x}}{x}\\\\
&=-2\lim_{x\to 0}\frac{d\left(1+x\right)^{1/x}}{dx}\\\\
&=-2\lim_{x\to 0}\left(1+x\right)^{1/x}\left(\frac{1}{x(x+1)}-\frac{\log(1+x)}{x^2}\right)\\\\
&=e
\end{align}$$

NOTE:
$$\begin{align}
\frac{1}{x(x+1)}-\frac{\log(1+x)}{x^2}&=\frac{1}{x(x+1)}-\frac{x-\frac12x^2+O(x^3)}{x^2}\\\\
&=\frac{-1}{x+1}+\frac12 +O(x)\\\\
&\to -\frac12
\end{align}$$
as $x\to 0$.
