Find the limit of the expression

[franktherabbit]

Homework Statement

$$\lim_{x\to\infty} \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}$$

Homework Equations

3. The Attempt at a Solution [/B]
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 don't think this is correct. My book gives $e^2$ as the solution. What do you think is wrong?

 

[Mark44]

Homework Statement

$$\lim_{x\to\infty} \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}$$

Homework Equations

3. The Attempt at a Solution [/B]
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 don't think this is correct. My book gives $e^2$ as the solution. What do you think is wrong?

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.

 

[franktherabbit]

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.

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?

 

[Mark44]

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?

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.

 

[Ray Vickson]

Homework Statement

$$\lim_{x\to\infty} \left(\frac{n^2+2n+1}{n^2+2n+3}\right)^{\frac{2n^2}{n+1}}$$

Homework Equations

3. The Attempt at a Solution [/B]
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 don't think this is correct. My book gives $e^2$ as the solution. What do you think is wrong?

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.