Evaluate $\lim_{x\to \infty} (\frac {x+1}{x-1})^x$ Evaluate $$\lim_{x\to \infty} \left(\frac {x+1}{x-1}\right)^x$$
My method:
$$\lim_{x\to \infty} \left(\frac {x+1}{x-1}\right)^x=\lim_{x\to \infty} \left(\frac {1+1/x}{1-1/x}\right)^x=1$$
Is that right?
 A: Not really, notice that 
$$\lim_{x \to \infty}\left( 1+\frac{y}{x}\right)^x=\exp(y)$$
Hence 
$$\lim_{x \to \infty} \left(\frac{1+\frac1x}{1-\frac1x} \right)^x=\frac{\lim_{x \to \infty}\left(1+\frac1x \right)^x}{\lim_{x \to \infty}\left(1-\frac1x \right)^x}=\frac{e}{e^{-1}}=e^2$$
A: Hint: No, you can’t treat the limits separately as you did in your final step:
$$\lim_{x \to \infty}\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)^x \color{red}{\neq \left(\frac{1+0}{1-0}\right)^x = 1}$$
Instead, note that
$$\lim_{x \to \infty}\left(\frac{x+1}{x-1}\right)^x = \lim_{x \to \infty}\left(1+\frac{2}{x-1}\right)^x = \lim_{x \to \infty}\left[\left(1+\frac{2}{x-1}\right)^{x-1}\right]^{\frac{x}{x-1}}$$
and make use of the standard limit of $e^x$.
A: Note that 


*

*$\lim_{x\to +\infty}\left(1 + \frac{a}{x} \right)^x = e^a$
So you have
\begin{eqnarray*}\left(\frac{x+1}{x-1}\right)^x
& = & \left(\frac{x\left(1+\frac{1}{x} \right)}{x\left(1-\frac{1}{x} \right)}\right)^x \\
& = & \frac{\left(1+\frac{1}{x}\right)^x}{\left(1-\frac{1}{x}\right)^x} \\
& \stackrel{x \to +\infty}{\longrightarrow} & \frac{e}{e^{-1}} = e^2
\end{eqnarray*}
