Epsilon proof of a sequence's limit - algebra issues

Question

I am attempting to prove $\lim_{k\to\infty}\dfrac{k^2}{k^2+2k+2}=1$. However, I am getting tripped up on the algebra. I believe I want to show that there exists some $N\in\mathbb{N}$ such that when $k\geq N$, $\left|\dfrac{k^2}{k^2+2k+2}-1\right|<\epsilon$ for arbitrary $\epsilon>0$. I set out to find a choice for $N$ by turning $1$ into $\dfrac{k^2+2k+2}{k^2+2k+2}$, but I am not sure where to go from $\left|\dfrac{-2k-2}{k^2+2k+2}\right|<\epsilon$. How can I rearrange this inequality to be in terms of $k$? Thank you!

Answer 1

Let's start with $\left|\dfrac{-2k-2}{k^2+2k+2}\right|=\left|\dfrac{2(k+1)}{k^2+2k+2}\right|<\left|\dfrac{2(k+1)}{(k+1)^2}\right|=\dfrac{2}{k+1}<\epsilon$. This inequality will hold if $k+1>\dfrac{2}{\epsilon}$ or $k>\dfrac{2-\epsilon}{\epsilon}$

Answer 2

You clearly get something that goes to zero. Try dividing by $k^2$ in both numerator and denominator: $$\left|\frac{-2k-2}{k^2+2k+2}\right|=\left|\frac{\frac{-2}k-\frac 2{k^2}}{1+\frac2k+\frac2{k^2}}\right|\lt\left|\frac2k +\frac2{k^2}\right |\lt\left|\frac2k+\frac2k \right |=\left|\frac4k\right |$$, for $k\gt1$.

So just choose $N$ such that $N\gt\frac 4{\epsilon}$.