Limit of $(\int\limits_0^n (1+\arctan^2x )\,dx )^ {\frac{1}{n}}$

Question

I need to find out this limit. Could someone help me? $$\lim_{n\rightarrow \infty} \Big(\int\limits_0^n (1+\arctan^2x )\,dx \Big)^ {\frac{1}{n}}$$ = ?

I have tried taking logarithm then calculating the integral of $arctan^{2}(x)$, it got worse, it seems to me that there is some shorter solution..

Answer 1

Your integrand is bounded as $$ 1\leq 1+\arctan^2x\leq 1+\frac{\pi^2}{4}. $$ Thus, your integral is between $n$ and $n(1+\pi^2/4)$. Taking the $n$th root and using the squeeze theorem for limits together with the facts that $$ \lim_{n\to+\infty}n^{1/n}=1 \qquad\text{and}\qquad \lim_{n\to+\infty}a^{1/n}=1 $$ should get you to the goal.

Answer 2

Hint: After taking the logarithm, try using l'hopital.