# finding and proving a limit using $\epsilon$

Use an $$\epsilon-N$$ argument to find and prove the $$\lim_{n\rightarrow\infty}\sqrt[n]{5+n^2}$$. Try some variations of your own.

I have to find and prove this limit using $$\epsilon-N$$ argument.

I think that the limit is 1, since the limit as n tends to infinity of the nth root of a polynomial(in this case it is $$5+n^2$$) is 1

Let's fix some $$\varepsilon > 0$$, and consider when $$|\sqrt[n]{5 + n^2} - 1| < \varepsilon$$. Note that $$\sqrt[n]{5 + n^2} - 1 > 0$$, so this is equivalent to finding where $$\sqrt[n]{5 + n^2} < 1 + \varepsilon \iff 5 + n^2 < (1 + \varepsilon)^n.$$ Both sides should tend to $$\infty$$, though the right tends eventually much faster. We can see this using the binomial theorem (for $$n \ge 2$$): \begin{align*} (1 + \varepsilon)^n &= 1 + n\varepsilon + \frac{1}{2}n(n+1)\varepsilon^2 + \ldots \\ &\ge 1 + n\varepsilon + \frac{1}{2}n(n+1)\varepsilon^2 \\ &\ge 1 + n\varepsilon + \frac{1}{2}n^2\varepsilon^2. \end{align*} So, if we can find $$N$$ such that $$n > N \implies 5 + n^2 < 1 + n\varepsilon + \frac{1}{2}n^2\varepsilon^2,$$ then we are done.
In fact, choose an $$N \ge \frac{\sqrt{2}}{\sqrt{\varepsilon}}$$, and you'll get that $$\frac{1}{2}n^2 \varepsilon^2 > n^2$$. If you choose $$N \ge \frac{4}{\varepsilon}$$, then $$1 + n \varepsilon > 5$$. Plus, of course, we must have $$n \ge 2$$ for our binomial theorem analysis to work. So, our choice of $$N$$ could be $$N = \max\left\{2, \frac{4}{\varepsilon}, \frac{\sqrt{2}}{\sqrt{\varepsilon}}\right\}.$$
• You should be able to make a better proof (IMHO) by getting just an $\epsilon$ and no roots of $\epsilon$. Feb 14, 2019 at 21:07
• @souadbouchahine The limit is 1, as the above proves. Note the $1$ in $|\sqrt[n]{n^2 + 5} - 1| < \varepsilon$. Feb 17, 2019 at 0:11