Prove the limit converges $\epsilon$ proof

Question

I believe there is an error in the solutions, and wanted to double check here.

I need to show the following sequence converges to the proposed limit.

$\lim \frac{1}{6n^2+1} = 0$

so we need to show that

$\forall \epsilon >0$ It holds that

$\lvert \frac{1}{6n^2+1} \rvert < \epsilon$

Thus we need to prove

$\frac{1}{\epsilon} < 6n^2+1$

$\frac{1-\epsilon}{\epsilon} < 6n^2$

$n > \sqrt{\frac{1-\epsilon}{\epsilon}}$

So we pick some $N \in \mathbb{N}$ such that $N > \sqrt{\frac{1-\epsilon}{\epsilon}}$

It then follow for $n \geq N$ implies $\frac{1}{6n^2+1} < \epsilon$

However the soln's disagree. they say we need pick an $N \in \mathbb{N}$ such that $N > \sqrt{\frac{1}{6 \epsilon}}$ which implies...

Isn't this incorrect?

Answer 1

See this

$$ \bigg| \frac{1}{6n^2+1} \bigg| < \frac{1}{6n^2} < \frac{1}{n} < \epsilon $$

Answer 2

We may write

$$\begin{align} \left| \frac{1}{6n^2+1} \right| &< \left| \frac{1}{6n^2} \right| \\ &<\epsilon \end{align}$$

whenever $n>\frac{1}{\sqrt{6\epsilon}}$