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?

Note that 


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*$\lim_{x\to +\infty}\left(1 + \frac{a}{x} \right)^x = e^a$
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*}
