Is it equation true - limit Is it true that :
$\lim_{ x\to\infty }  \left( 1+\frac{f \left( x \right) }{x} \right) ^x = \exp \left(  \lim_{ x\to\infty } f \left( x \right)   \right)$ ?
Assumption is that limit of $\lim_{ x\to\infty } f(x)$ exists.
 A: \begin{eqnarray}
\lim_{x\to\infty}\left(1+\frac{f(x)}x\right)^x
&=&\lim_{x\to\infty}\exp \left(x\log \left(1+\frac{f(x)}x \right) \right)=\exp \left(\lim_{x\to\infty}x\log \left(1+\frac{f(x)}x\right)\right)\\ \ \\
&=&\exp \left(\lim_{x\to\infty}x \left(\frac{f(x)}x+o\left(\frac1{x^2}\right)\right)\right)
=\exp \left(\lim_{x\to\infty}f(x)+o\left(\frac1{x}\right)\right)\\ \ \\
&=&\exp \left(\lim_{x\to\infty}f(x)\right).
\end{eqnarray}
The limit exchange in the second equality is justified by the fact that the exponential is continuous and that the expression inside converges.
A: Since there exist $\lim\limits_{x\to 0}f(x)=l$ we have that $f$ is bounded in the neighborhood of $\infty$. So $\lim\limits_{x\to\infty}\frac{f(x)}{x}=0$ and we can write
$$
\lim\limits_{x\to\infty}\left(1+\frac{f(x)}{x}\right)^x=
\lim\limits_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{f(x)}=
\lim\limits_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{\lim\limits_{x\to\infty}f(x)}=
e^l
$$
