Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$ Facing difficulty finding limit
$$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$
For starters I have trouble simplifying it
Which method would help in finding this limit?

If you know that
$$\lim_{x\to\infty}\left(1 + \frac{a}{x}\right)^x = e^{a},$$
so that
$$\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x = e^{-1},$$
then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product,
$$\begin{align*}
\left(\frac{x}{x-1}\right)^{2x+1} &= \left(\left(\frac{x}{x-1}\right)^x\right)^2\left(\frac{x}{x-1}\right)\\
&= \left(\frac{1}{\left(\frac{x-1}{x}\right)^x}\right)^2\left(\frac{x}{x-1}\right)\\
&= \left(\frac{1}{\left(1 - \frac{1}{x}\right)^x}\right)^2\left(\frac{x}{x-1}\right).
\end{align*}$$
Then use limit laws to compute it.
