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I'm trying to find the following limits using Dominated Convergence Theorem, but can't seem to find a dominating function. Any guidance would be greatly appreciated!

$\lim\limits _{n\rightarrow\infty}\int_{0}^{1}\frac{1+n^2x^2}{(1+x)^{n}}dx$

$\lim_{n\rightarrow\infty}\int_{0}^{\infty}ne^{-nx}\sin(1/x)dx$

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For the second problem, it is straightforward to find a (non-negative) dominating function on $[1,\infty)$.

We have $\sin(1/x) > 0$ for $x \in [1,\infty)$ and

$$e^{nx} \geqslant 1 + nx \implies |ne^{-nx}\sin(1/x)| \leqslant\frac{n}{1+nx}\sin(1/x)\leqslant \frac1{x}\sin(1/x),$$

Making the change of variables $u = 1/x$, we see that this choice of dominating function is integrable over $[1,\infty)$:

$$\int_1^{\infty}\frac1{x} \sin(1/x) \, dx= \int_1^{\infty}\frac{ \sin u}{u} \, du < \infty.$$

Since $\displaystyle \lim_{n \to \infty} ne^{-nx}\sin(1/x) = 0$ for all $x > 0$, by the DCT it follows that

$$\lim_{n \to \infty}\int_1^{\infty}ne^{-nx}\sin(1/x) \, dx=0.$$

Alternatively, this holds true for integrals over any interval $[a,\infty)$ with $a > 0$ -- following from the uniform convergence of the sequence of integrands. It is also true with $a= 0$, but it is not likely that this can be demonstrated using the DCT.

For example, there is no dominating integrable function for $ne^{-nx}$, since $\displaystyle \lim_{n \to \infty} ne^{-nx}= 0$ and

$$\lim_{n \to \infty}\int_0^{\infty}ne^{-nx}\, dx = 1.$$

The limit of the integral over $(0,\infty)$ is shown to be $0$ using another approach in: https://math.stackexchange.com/a/59567/148510

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  • $\begingroup$ Thank you! Makes complete sense. Knowing that first inequality really sets the whole thing up. I was trying to think of something like that, but was not aware that $e^{nx}\geq1+nx$ for $x\in(0,\infty)$ $\endgroup$ – Math244 Jul 14 '15 at 18:16
  • $\begingroup$ Your welcome. For the first problem show that $\lim_{n \to \infty}f_n(x) = 0$ for $x \in (0,1]$. Then $f_n(x) \leqslant 1$ for large n. $\endgroup$ – RRL Jul 14 '15 at 18:22
  • $\begingroup$ @RRL It seems there is still a bit of work to do to show that $f_n(x)\le 1$ for "n large." How does one know, without analysis, that for a fixed $n$ we cannot find a value of $x$ for which the inequality is violated? $\endgroup$ – Mark Viola Jul 14 '15 at 22:15
  • $\begingroup$ @Dr. MV: For sure. Just offered this as a hint. The local maximum of $f_n(x)$ inside $(0,1]$ falls below $1$ for $n \geqslant 5$ and $f_n(0) = 1$. $\endgroup$ – RRL Jul 14 '15 at 22:42
  • $\begingroup$ $f=1$ is actually is a global maximum for all $n\ge1$, I believe. There is one local maximum and one local minimum that are less than $1$. $\endgroup$ – Mark Viola Jul 14 '15 at 22:52

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