# Thread: Hard limit problem

1. ## Hard limit problem

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

Without the n^2+2 in exponent would be easy. But how do we get rid of that?

Sorry if it's a stupid question, but in school we didn't solve these kind of problems.

The result should be e^2 or something like that.

2. Originally Posted by Nforce
$\displaystyle \displaystyle\lim_{n\to\infty}(\frac{n^2+3}{n^2+1} )^{n^{2}+2}$

Without the n^2+2 in exponent would be easy. But how do we get rid of that?

Sorry if it's a stupid question, but in school we didn't solve these kind of problems.

The result should be e^2 or something like that.
$\displaystyle \displaystyle\lim_{n\to\infty}\left[\left(\frac{n^2+3}{n^2+1}\right)^{n^2}\cdot\left(\ frac{n^2+3}{n^2+1}\right)^2\right]$

$\displaystyle \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n ^2+1}\right)^{n^2}\cdot\lim_{n\to\infty}\left(\fra c{n^2+3}{n^2+1}\right)^2$

Does this help you?

3. Originally Posted by Nforce
$\displaystyle \displaystyle\lim_{n\to\infty}(\frac{n^2+3}{n^2+1} )^{n^{2}+2}$

Without the n^2+2 in exponent would be easy. But how do we get rid of that?

Sorry if it's a stupid question, but in school we didn't solve these kind of problems.

The result should be e^2 or something like that.
$\displaystyle \displaystyle\left(\frac{n^2+3}{n^2+1}\right)^{n^2 +2}=\left(\frac{(n^2+1)+2}{n^2+1}\right)^{(n^2+1)+ 1$

$\displaystyle =\displaystyle\left[\left(1+\frac{2}{n^2+1}\right)^{\frac{n^2+1}{2}}\r ight]^2\left(1+\frac{2}{n^2+1}\right)$

Now you can use

$\displaystyle \displaystyle\ e=\lim_{x\ \to \infty}\left(1+\frac{1}{x}\right)^x$

4. Hello, Nforce!

Are you allowed to use L'Hopital?

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

$\displaystyle \text{We have: }\;y \;=\;\left(\dfrac{n^2+3}{n^2+1}\right)^{n^2+2}\;\; \to\;\;1^{\infty}$

$\displaystyle \text{Take logs: }\;\ln y \;=\;\ln\left(\dfrac{n^2+3}{n^2+1}\right)^{n^2+2} \;=\;(n^2+2)\ln\left(\dfrac{n^2+3}{n^2+1}\right) \;\;\to\;\;\infty \cdot 0$

. . . . . . . . .$\displaystyle \ln y \;=\;\dfrac{\ln(n^2+3) - \ln(n^2+1)}{(n^2+2)^{-1}} \;\;\to\;\;\dfrac{\infty - \infty}{0}$

$\displaystyle \text{Apply l'Hopital: }\;\ln y \;=\;\dfrac{\dfrac{2n}{n^2+3} - \dfrac{2n}{n^2+1}}{-\dfrac{2n}{(n+2)^2}} \;=\;\dfrac{2n\left(\dfrac{n^2+1 - n^2-3}{(n^2+3)(n^2+1)}\right)}{-\dfrac{2n}{(n^2+2)^2}}$

. . . . . . . . . . . . . . $\displaystyle \ln y \;=\;\dfrac{\dfrac{2n(-2)}{(n^2+3)(n^2+1)}}{\dfrac{-2n}{(n^2+2)^2}} \;=\;\dfrac{2(n^2+2)^2}{(n^2+3)(n^2+1)}$

Divide numerator and denominator by $\displaystyle n^4$

. . . . $\displaystyle \ln y \;=\; \dfrac{2\left(1 + \frac{2}{n^2}\right)^2} {\left(1 + \frac{3}{n^2}\right)\left(1 + \frac{1}{n^2}\right)}$

Then: .$\displaystyle \displaystyle \lim_{x\to\infty} \ln y \;=\;\lim_{x\to\infty} \dfrac{2\left(1 + \frac{2}{n^2}\right)^2} {\left(1 + \frac{3}{n^2}\right)\left(1 + \frac{1}{n^2}\right)} \;=\;\frac{2(1+0)^2}{(1+0)(1+0)} \;=\;2$

$\displaystyle \displaystyle \text{Since }\lim_{x\to\infty}\ln y \;=\;2,\,\text{ then: }\lim_{x\to\infty} y \;=\;e^2$

5. Yes we have used the L'hopital. But the thing is that we never solved limits with "n" in exponent. I feel sorry for you soroban, thanks for help but I don't understand the whole process, what does that arrow mean? and why did you write 1^infinity.

I think this is something that I understand more. Just simple math rules.
$\displaystyle \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n ^2+1}\right)^{n^2}\cdot\lim_{n\to\infty}\left(\fra c{n^2+3}{n^2+1}\right)^2$

But I have some question tho.

Could I rewrite that into this, and solve it

$\displaystyle \displaystyle\lim_{n\to\infty}\left(\frac{n^2+3}{n ^2+1}\right)^{n^2}\cdot\left(\lim_{n\to\infty}\fra c{n^2+3}{n^2+1}\right)^2$

But the first term has still n in the exponent. Oh man, can I write it into a fraction someway?

6. You could simplify the process if you are unfamiliar with the limit expression for $\displaystyle e^x$

$\displaystyle y=n^2+2$

$\displaystyle \displaystyle\left(\frac{n^2+3}{n^2+1}\right)^{n^2 +2}=\left(\frac{y+1}{y-1}\right)^y$

Take natural logs

$\displaystyle \displaystyle\ lnL=\lim_{y \to \infty}ln\left(\frac{y+1}{y-1}\right)^y=\lim_{y \to \infty}yln\left(\frac{y+1}{y-1}\right)=\lim_{y \to \infty}\frac{ln\left(\frac{y+1}{y-1}\right)}{\left(\frac{1}{y}\right)}$

which $\displaystyle \rightarrow\frac{0}{0}$

Use L'Hopital's Rule, differentiate with the chain rule (and the quotient rule)....

$\displaystyle \displaystyle\lim_{y \to \infty}\frac{\frac{y-1}{y+1}\left[\frac{(y-1)(1)-(y+1)(1)}{(y-1)^2}\right]}{\left(-\frac{1}{y^2}\right)}$

and

$\displaystyle \displaystyle\frac{y-1}{y+1}\left[\frac{-2}{(y-1)^2}\right]\left(-y^2}\right)$

$\displaystyle =\displaystyle\frac{(y-1)2y^2}{(y+1)(y-1)^2}=\frac{2y^2}{(y+1)(y-1)}=\frac{2y^2}{y^2-1}$

$\displaystyle \displaystyle\lim_{y \to \infty}\frac{2y^2}{y^2-1}=2$

$\displaystyle L=e^2$

7. I take y_n = n^2 + 1. The piece inside the exponent becomes

$\displaystyle \left(1 + \frac{2}{y_n}\right)$

so the full expression is

$\displaystyle \left(1 + \frac{2}{y_n}\right)^{y_n} \cdot \left(1 + \frac{2}{y_n}\right)$

Let y_n go to infinity and get e^2.

This is the same as Archie's solution, but prettier to look at because we don't have a bunch of ugly things floating around

8. ## Need Help

Can anyone tel me?
what is the value of tanh(x)/x when x tends to infinity and what is the value of sinh(x)/x when x tends to infinity?

9. Originally Posted by skullface
Can anyone tel me?
what is the value of tanh(x)/x when x tends to infinity and what is the value of sinh(x)/x when x tends to infinity?
start a new problem w/ a new thread ... don't tag, please.

10. Respect!, Now I understand, thanks a lot.