How to calculate $\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$ I know from an online calculator http://www.numberempire.com/derivatives.php that 
$\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$.  How do you calculate this step by step?
 A: For ease rewrite as 
$$
\lim_{n\rightarrow \infty}\left(1+\frac{2}{x}\right)^x.
$$
First we compute 
$$
\lim_{n\to\infty}\ln\left(\left(1+\frac{2}{x}\right)^x\right).
$$
Using laws of logarithms we get 
$$
\lim_{n\to\infty}x\ln\left(1+\frac{2}{x}\right)  =\lim_{n\to\infty}\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}.
$$
We are now in a position to apply L'Hopital's Rule.  Taking derivatives gives
$$
\lim_{n\to\infty} \frac{\left(\frac{1}{1+\frac{2}{x}}\right)\left(-\frac{2}{x^2}\right)}{-\frac{1}{x^2}}=\lim_{n\to\infty}\frac{2}{1+\frac{2}{x}}=2.
$$
Now, your limit is 
$$
\lim_{n\to\infty}e^{\ln\left(\left(1+\frac{2}{x}\right)^x\right)}=e^2
$$
A: Rewriting the function whose limit is taken $$\left(1 + \frac{2}{x}\right)^x,$$ we immediately recognize the definition of $e^2$.
A: It's easy to obtain 
$\lim _{x\rightarrow \infty }\left( \dfrac {x+2} {x}\right) ^{x}$ =$\lim _{x\rightarrow \infty }\left( \dfrac {x} {x}+\dfrac {2} {x}\right) ^{x}$
and  it's easy for $\lim _{x\rightarrow \infty }\left( \dfrac {x} {x}+\dfrac {2} {x}\right) ^{x}$=$\lim _{x\rightarrow \infty }\left( 1+\dfrac {2} {x}\right) ^{x}$
do some conversion
$\lim _{x\rightarrow \infty }\left[ \left( 1+\dfrac {2} {x}\right) ^{\dfrac {x} {2}}\right] ^{2}$=$\lim _{\dfrac {x} {2}\rightarrow \infty }\left[ \left( 1+\dfrac {2} {x}\right) ^{\dfrac {x} {2}}\right] ^{2}$=$\lim {t\rightarrow \infty }\left[ \left( 1+\dfrac {1} {t}\right) ^{t}\right] ^{2}$($\dfrac {x} {2}=t$)=$e^{2}$(use $\lim {t\rightarrow \infty }$$\left( 1+\dfrac {1} {t}\right) ^{t}$=$e$)
A: This is more or less by definition of $e$, depending on which definition you use. Do you know the definition $e=\lim_{n\rightarrow\infty} (1+\frac{1}{n})^n$? Then set $x=2n$ and you are done.
A: Observe that
$$\left( \dfrac{x+2}{x}\right) ^{x}=\left( \left( 1+\dfrac{1}{x/2}
\right) ^{x/2}\right) ^{2}.$$
Hence, by the definition of $e$
$$e=\lim_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}$$
we have
$$\begin{eqnarray*}
\lim_{x\rightarrow \infty }\left( \dfrac{x+2}{x}\right) ^{x}
&=&\lim_{x\rightarrow \infty }\left( \left( 1+\frac{1}{x/2}\right) ^{x/2}\right) ^{2} \\
&=&\left( \lim_{x\rightarrow \infty }\left( 1+\frac{1}{x/2}\right) ^{%
x/2}\right) ^{2} \\
&=&\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right)
^{2} \\
&=&e^{2}
\end{eqnarray*}$$

For instance, by the same argument, for $k\in\mathbb{N}$, we deduce that
$$\lim_{x\rightarrow \infty }\left( \dfrac{x+k}{x}\right) ^{x}=\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right)
^{k}=e^{k}$$
A: Find a step-by-step derivation here (and not only for this special problem):
http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x)%5Ex+x-%3Eoo
(if the link doesn't work copy and paste the whole line)
...or go to http://www.wolframalpha.com directly and type:
"limit ((x+2)/x)^x x->oo"
Click on "Show steps" - Done! ;-)
