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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..

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    $\begingroup$ What about your thoughts on the problem? Did you try to take logarithm of both sides, and use l'Hopital rule? Please include it in your question $\endgroup$ – Jakobian Oct 10 '18 at 18:17
  • $\begingroup$ Do you know the limit of $n^{1/n}$ as $n\to+\infty$? $\endgroup$ – mickep Oct 10 '18 at 18:27
  • $\begingroup$ @Jakobian, 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.. $\endgroup$ – Emathke Oct 10 '18 at 18:31
  • $\begingroup$ @Emathke I actually said "Please include it in your question" to avoid you writing it to me in the comments $\endgroup$ – Jakobian Oct 10 '18 at 18:34
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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.

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

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