Find $\lim_{n\rightarrow\infty}\sqrt[n]{5+n^2}$ using $\varepsilon-N$ language

Question

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

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

Answer 1

Let $\sqrt[n]{5+n^2}=s_n$ now solve the limit: $\lim_{n\rightarrow\infty}log(s_n)=\lim_{n\rightarrow\infty}\frac{log(5+n^2)}{n}=0$=>$\lim_{n\rightarrow\infty}\sqrt[n]{5+n^2}=1$; The step 2: fix $\epsilon_0>0$, we must show $0<\sqrt[n]{5+n^2}<\epsilon_0+1$ we know that the equation:$$log(\epsilon_0)=\frac{log(5+x^2)}{x}\hspace{0.5cm}(*)$$ have a solution in $\mathbb R^* $ for all $\epsilon_0$. Let the solution be $x(\epsilon_0)$. Then, because equation $(*)$ is equivalent with equation $(5+x^2)^{\frac{1}{x}}=\epsilon_0$ and function $f(x)=(5+x^2)^{\frac{1}{x}}$ decreases, we obtain: for all $n>=[x(\epsilon_0)]+1$ we have $(5+n^2)^{\frac{1}{n}}<\epsilon_0<\ \epsilon_0+1$. where:$\hspace{0.5cm}[x(\epsilon_0)]$ is integer part of $x(\epsilon_0)$

Answer 2

Let $a_n=\sqrt[n]{5+n^2}-1$ and then, for $n\ge3$, $$ 2n^2\ge5+n^2=(a_n+1)^n\ge \binom{n}{3}a_n^3. $$ So $$ 0\le a_n\le \sqrt[3]{\frac{12n^2}{n(n-1)(n-2)}}\le3\sqrt[3]\frac{1}{n-2}. $$ For $\forall\epsilon>0$, define $$ N=3+\bigg[\frac{3}{\epsilon}\bigg]^3 $$ and then, if $n\ge N$, $$ 0\le a_n<\epsilon $$ or $$ |\sqrt[n]{5+n^2}-1|<\epsilon. $$