Prove with the help of the $\varepsilon$-$n_0$ definition, that $\frac 1 4$ is the limit of the sequence $a_n=\frac{n^2-1}{4n^2+1}$

Question

Prove with the help of the $\varepsilon$-$n_0$ definition, that $\frac 1 4$ is the limit of the number-sequence $a_n=\frac{n^2-1}{4n^2+1}$

Let $\varepsilon>0$. We need to show, that $\frac 1 4 - \varepsilon < \frac{n^2-1}{4n^2+1} < \frac 1 4 +\varepsilon$

"Upper bound": $\frac{n^2-1}{4n^2+1}\leq \frac{n^2+\frac 1 4}{4n^2+1}=\frac 1 4 < \frac 1 4 +\varepsilon$ if $\varepsilon > 0$.

"Lower bound": $\frac{n^2-1}{4n^2+1}\geq \frac{n^2-1}{4(n+1)^2}=\frac{(n-1)(n+1)}{4((n+1)(n+1))}=\frac{n-1}{4(n+1)}$ how to go on now? I can't seem to find a way to simplify this more. Can you help me a bit?

Answer 1

We have $$\left|\frac{n^2-1}{4n^2+1}-\frac{1}{4}\right|=\left|\frac{4n^2-4-4n^2-1}{4(4n^2+1)}\right|=\frac{5}{4(4n^2+1)}=\frac{5}{4(4n^2+1)}<\frac{5}{16n^2}$$ I hope you can finish now!