# Find the limit of the expression

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1. Jan 1, 2017

### franktherabbit

1. The problem statement, all variables and given/known data
$$\lim_{x\to\infty} \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}$$

2. Relevant equations
3. The attempt at a solution

I tried
$\lim_{x\to\infty} \left(\frac{n^2+2n+3-2}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}=$
$\lim_{x\to\infty} \left(1+\frac{-2}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}=$
$\lim_{x\to\infty} \left(1 +\frac{-2}{n^2+2n+3}\right)^{\frac{n^2+2n+3}{-2}\frac{-2}{n^2+2n+3}\frac{2n^2}{n+1}}=$
$e^{\lim_{x\to\infty}\frac{-4n^2}{(n^2+2n+3)(n+1)}}=1$
and i get 1 but i dont think this is correct. My book gives $e^2$ as the solution. What do you think is wrong?

2. Jan 1, 2017

### Staff: Mentor

Your limit is of the indeterminate form $[1^{\infty}]$. The usual way to deal with this type of problem is to let $u = \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}$, and then take the natural log of both sides. Then take the limit, keeping in mind that you can usually switch the order of the limit and ln operations.

3. Jan 1, 2017

### franktherabbit

So, $\ln u=\ln \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}$
$\ln u=\frac{2n^2}{n+1}\ln \left(\frac{n^2+2n+1}{n^2+2n+3}\right)$
$\lim_{n\to\infty}\ln u=\lim_{n\to\infty}\frac{2n^2}{n+1}\ln \left(\frac{n^2+2n+1}{n^2+2n+3}\right)$
$\lim_{n\to\infty}\ln u=\lim_{n\to\infty}\frac{2n^2}{n+1}\ln \left(\lim_{n\to\infty}\frac{n^2+2n+1}{n^2+2n+3}\right)$
Is this what you meant? I still get infinity at the first part, how to deal with this?

4. Jan 1, 2017

### Staff: Mentor

The first two steps look OK, but not after that. In the 2nd equation, write the righthand side as $\frac{\ln \left(\frac{n^2 + 2n + 1}{n^2 + 2n + 3}\right)}{\frac{n + 1}{2n^2}}$. This is of the indeterminate form $[\frac 0 0 ]$, so L'Hopital's Rule applies.

5. Jan 1, 2017

### Ray Vickson

Your result is correct:
$$\lim_{n \to \infty} \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}=1$$
Note: the limit is for $n \to \infty$, not some mythical $x$ going to $\infty$.

This is easy to check:
$$\ln \frac{n^2+2n+1}{n^2+2n+3} = -\frac{2}{n^2} + O\left(\frac{1}{n^3}\right),$$
so
$$\frac{2n^2}{n+1} \ln \frac{n^2+2n+1}{n^2+2n+3} = -\frac{4}{n} + O \left( \frac{1}{n^2} \right) \to 0.$$
Since the logarithm goes to 0 the function itself goes to 1.

Also: when the problem is submitted to Maple, the limit is given as 1.

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