# 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?

• This is already solved, but another useful hint: $$\frac{x}{x-1} = \frac{x-1+1}{x-1} = 1+\frac{1}{x-1}.$$ Dec 11 '11 at 5:11

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.

• I love this answer! Very elegant! Dec 11 '11 at 1:08
• An initial substitution of y=x-1 makes it a lot easier.
– tzs
Dec 11 '11 at 4:45
• @tzs maybe you mean $y = x+1$. Check out my answer. Dec 11 '11 at 5:00

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.

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

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

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

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