Vtyjkyuk

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

You clearly get something that goes to zero. Try dividing by $k^2$ in both numerator and denominator: $$left|frac-2k-2k^2+2k+2right|=left|fracfrac-2k-frac 2k^21+frac2k+frac2k^2right|ltleft|frac2k +frac2k^2right |ltleft|frac2k+frac2k right |=left|frac4kright |$$, for $kgt1$.

So just choose $N$ such that $Ngtfrac 4epsilon$.

Let's start with $left|dfrac-2k-2k^2+2k+2right|=left|dfrac2(k+1)k^2+2k+2right|<left|dfrac2(k+1)(k+1)^2right|=dfrac2k+1<epsilon$. This inequality will hold if $k+1>dfrac2epsilon$ or $k>dfrac2-epsilonepsilon$