Compute this limit using the relation between Riemann and Lebesgue integral

Question

Can anyone help me compute the following limit? $$ \lim_{n \rightarrow \infty}{ \int_0^{\infty}\left(\frac{nx^{1/n}}{ne^x+\sin(nx)}\right)dx}$$

Using the relation between Riemann and Lebesgue integral and Lebesgue’s Dominated Convergence Theorem.

Answer 1

Hint. Note that for $x>0$, $$\lim_{n\to +\infty}\frac{nx^{1/n}}{ne^x+\sin(nx)}= \lim_{n\to +\infty}\frac{x^{1/n}}{e^x+\frac{\sin(nx)}{n}}=e^{-x}.$$ Moreover for $n\geq 2$, $$0\leq\frac{nx^{1/n}}{ne^x+\sin(nx)}=\frac{x^{1/n}}{e^x+\frac{\sin(nx)}{n}}\leq \frac{\max(1,x)}{e^x-\frac{1}{2}}.$$

Answer 2

For $x>0$, we have $f_n(x) = { \sqrt[n]{x} \over e^x + x { \sin (nx) \over nx} } $, and since $ |\operatorname{sinc} x|\le 1$, we have $f_n(x) \le { \sqrt[n]{x} \over e^x - x } \le { \max(1,x) \over 1+{1 \over 2!} x^2 +{1 \over 3!} x^3 }$, and since the latter is integrable, we can use the dominated convergence theorem to swap the integration and limit to get $\int f_n \to \int f$, where $f(x) = e^{-x}$. The latter integral is straightforward to evaluate.