# 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?

• Is $\lim_{x\to\infty}(1+1/x)^x = 1$? – Ennar Jan 9 '19 at 13:32

Not really, notice that

$$\lim_{x \to \infty}\left( 1+\frac{y}{x}\right)^x=\exp(y)$$

Hence

$$\lim_{x \to \infty} \left(\frac{1+\frac1x}{1-\frac1x} \right)^x=\frac{\lim_{x \to \infty}\left(1+\frac1x \right)^x}{\lim_{x \to \infty}\left(1-\frac1x \right)^x}=\frac{e}{e^{-1}}=e^2$$

Hint: No, you canâ€™t treat the limits separately as you did in your final step:

$$\lim_{x \to \infty}\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)^x \color{red}{\neq \left(\frac{1+0}{1-0}\right)^x = 1}$$

$$\lim_{x \to \infty}\left(\frac{x+1}{x-1}\right)^x = \lim_{x \to \infty}\left(1+\frac{2}{x-1}\right)^x = \lim_{x \to \infty}\left[\left(1+\frac{2}{x-1}\right)^{x-1}\right]^{\frac{x}{x-1}}$$
and make use of the standard limit of $$e^x$$.
• $$\lim_{x\to +\infty}\left(1 + \frac{a}{x} \right)^x = e^a$$
So you have $$\begin{eqnarray*}\left(\frac{x+1}{x-1}\right)^x & = & \left(\frac{x\left(1+\frac{1}{x} \right)}{x\left(1-\frac{1}{x} \right)}\right)^x \\ & = & \frac{\left(1+\frac{1}{x}\right)^x}{\left(1-\frac{1}{x}\right)^x} \\ & \stackrel{x \to +\infty}{\longrightarrow} & \frac{e}{e^{-1}} = e^2 \end{eqnarray*}$$