Here's a trickier one.
Prove $\displaystyle \lim_{n\to\infty}n^2e^{-n}=0$
Hint: Rewrite $\displaystyle n^2e^{-n}$ as $\displaystyle e^{\ln(n^2)}e^{-n}=e^{2\ln(n)-n}$ and then try to put an upper bound on $\displaystyle 2\ln(n)$ (that's easy to work with).
Note that if you follow the hint, your value for $\displaystyle N$ will look something like $\displaystyle N=\max\{c, f(\epsilon)\}$ for some number $\displaystyle c$ and some expression depending on $\displaystyle \epsilon$.
I hope what I wrote wasn't too confusing.
Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at
$\displaystyle \frac{1}{\sqrt{n^2+1}+n} < \epsilon$
$\displaystyle 1 < {\epsilon}{{\sqrt{n^2+1}+n}}$
Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes
Could I even do
$\displaystyle {\sqrt{n^2+1}+n} > \frac{1}{\epsilon}$
Edit: I must continue so my erroneous ways can be corrected, even if it hurts, not literally.
$\displaystyle {n^2+1+n^2} > ({\frac{1}{\epsilon}})^2$
$\displaystyle {2n^2+1} > ({\frac{1}{\epsilon}})^2$
$\displaystyle {2n^2} > ({\frac{1}{\epsilon}})^2 - 1$
Stuck
What Drexel did was what I had in mind for the solution. If you do it this, way, you're going to get something like:
$\displaystyle \sqrt{n^2+1}+n>\frac{1}{\epsilon}\implies 2n^2+1{\color{red}+2n\sqrt{n^2+1}}>(1/\epsilon)^2\implies$ $\displaystyle 2n\sqrt{n^2+1}>(1/\epsilon)^2-1-2n^2 \implies 4n^2(n^2+1)>[(1/\epsilon)^2-1-2n^2]^2$
and solving that equation for $\displaystyle n$ will give you a nasty quadratic.
All you have to realize is that saying $\displaystyle \frac{1}{\text{whatever}}<\frac{1}{\text{whatever else}}$ assuming positivity is that $\displaystyle \text{whatever}>\text{whatever else}$. Now just note that $\displaystyle n^2+1>n^2\implies \sqrt{n^2+1}>\sqrt{n^2}=n\implies \sqrt{n^2+1}+n>2n\implies \sqrt{n^2+1}+n>n$