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Let $f_n (x) = \frac{nx^{1/n}}{ne^x + \sin(nx)}.$ The question is: with the dominated convergence theorem find the limit $ \lim_{n\to\infty} \int_0^\infty f_n (x) dx. $ So I need to find an integrable function $g$ such that $|f_n| \leq g$ for all $n\in \mathbf N$. I tried $ \frac{nx^{1/n}}{ne^x + \sin(nx)} = \frac{x^{1/n}}{e^x + \sin(nx)/n} \leq \frac{x^{1/n}}{e^x - 1} \leq \frac{x^{1/n}}{x}. $ But I can't get rid of that $n$. Can anyone give me a hint?

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Consider $n\geqslant2$, then $ |f_n(x)|\leqslant\frac{x^{1/n}}{\mathrm e^x-\frac12}\leqslant\frac{1+x}{\frac12\mathrm e^x}=g(x). $

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We have \begin{align} \left| \frac{nx^{1/n}}{ne^x + \sin(nx)} \right|= & \frac{|x^{1/n}|}{|e^x + \sin(nx)/n|} & \\ \leq & \frac{\max\{1,x\}}{|e^x + \sin(nx)/n|} & \mbox{by } |x^{1/n}|\leq \max\{1,x\} \\ \leq & \frac{\max\{1,x\}}{|e^x -\epsilon |} & \mbox{if } |e^x + \sin(nx)/n|\geq |e^x -\epsilon| \\ \end{align} Note that for all $\epsilon\in(0,1)$ existis $N$ any sufficiently large such that $ \left|\frac{1}{n}\sin(nx)\right|<\epsilon \implies |e^x + \sin(nx)/n|\geq |e^x -\epsilon| $ if $n>N$.

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    $x$ is non-negative, so $e^x \geq 1$ and of course $\sin(nx)/n \geq -1$, and hence $e^x + \sin(nx)/n \geq 0$. The absolute values are not needed, I think.2012-12-18