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?
 A: First, try finding the limit of its logarithm.  If you write $(2x+1)\cdot(\text{something}) = \frac{\text{something}}{1/(2x+1)}$, then L'Hopital's rule should do it.  Then take the antilogarithm of that and you've got it.
A: This is going to be very similar to what Arturo suggested but it has the benefit of arriving at the answer quicker. Using a substitution $x \mapsto y+1$ we can write the function as
$$\left(\frac{x}{x-1}\right)^{2x+1} = \left(1+\frac{1}{y}\right)^{2y+3} =\left(\left(1+\frac{1}{y}\right)^y\right)^2\left(1+\frac{1}{y}\right)^{3}$$
Finding the limit of this one should be easy
A: $$
\begin{eqnarray}
\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}=\lim \limits_{x\to \infty}\left(\frac{x-1+1}{x-1}\right)^{2x+1}
=\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{2x+1}\\= \lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}
 =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}=e^{\lim \limits_{x\to \infty}\frac{2x+1}{x-1}}=e^2
\end{eqnarray}
$$
A: 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.
A: $\lim_{x\to\infty}(\frac{x}{x-1})^{2x+1}=\lim_{x\to\infty}(\frac{x-1+1}{x-1})^{2x+1}=\lim_{x\to\infty}(1+\frac{1}{x-1})^{2x+1}=e^{\lim_{x\to\infty}\frac{2x+1}{x-1}}=e^2$.
A: Limits of the type f(x)^g(x) where f(x) tends to 1 and g(x) to infinity i.e 1^infinity
can be done as e^((f(x)-1).g(x))
if you do that you get e^2
