Limit of sequence of integrals

Question

$$\lim_{n\rightarrow \infty} \int _0^\infty \frac{(1-e^{-x})^n}{1+x^2}dx.$$

this is less then $$\lim_{n\rightarrow \infty} \int _0^\infty \frac{1}{1+x^2}dx.$$ so the limit should exist by DCT. But how do I calculate: $$ \int _0^\infty \lim_{n\rightarrow \infty} \frac{(1-e^{-x})^n}{1+x^2}dx$$

Answer 1

Following the argument of Bertrand R and DonAntonio, for all $x > 0$ we have $0 < (1 - e^{-x}) < 1$ and therefore $\lim_{n\to\infty}(1 - e^{-x})^n = 0$. So $$ \lim_{n\to\infty}\int_0^\infty \frac{(1 - e^{-x})^n}{1 + x^2} \mathrm dx \overset{\text{DCT}}{=}\int_0^\infty\lim_{n\to\infty}\frac{(1 - e^{-x})^n}{1 + x^2} \mathrm dx = \int_0^\infty 0 \;\mathrm dx = 0. $$