Limits of a definite integrals (Demidovich)

I'm being unable to solve this exercise from the Demidovich's book where you need to evaluate limits involving definite integrals where the interval of integration is defined as a function of x.

How do I approach this kind of problems? Here I list one specific problem from Demidovich's book: $$\lim_{x\to+\infty} \frac{\int_{0}^{x} (\arctan (t))^2 dt}{\sqrt{x^2+1}}$$

Could you please demonstrate the approach on the problem above?

1 Answer

The limit in question is an indeterminate form (in this case $\frac\infty\infty$), so you would apply L'Hopital's rule. To differentiate the numerator, use the Fundamental Theorem of Calculus.

• I applied L'Hospital's rule and I'm left with $\lim_{x\to+\infty} \frac{(\arctan(x))^2}{\frac{x}{\sqrt{x^2+1}}}$. How do I proceed? – Gogis May 4 '16 at 22:42
• The denominator tends to $1$ because you can factor out $x$ from both top and bottom. As for the numerator, consult a plot of the arctan function to see that $\arctan x$ tends to $\pi/2$ as $x\to\infty$ – grand_chat May 4 '16 at 22:50
• Yes, I see. Everything is clear to me now. Thanks. – Gogis May 4 '16 at 22:52