Need help finding limit $\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

$$ \begin{eqnarray} \lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}=\lim \limits_{x\to \infty}\left(\frac{x-1+1}{x-1}\right)^{2x+1} =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{2x+1}\\= \lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}} =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}=e^{\lim \limits_{x\to \infty}\frac{2x+1}{x-1}}=e^2 \end{eqnarray} $$

Answer 5

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