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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!

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

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

Math244
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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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RRL
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  • 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 $x\in(0,\infty)$ – Math244 Jul 14 '15 at 18:16
  • 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. – RRL Jul 14 '15 at 18:22
  • @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? – Mark Viola Jul 14 '15 at 22:15
  • @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 $f_n(0) = 1$. – RRL Jul 14 '15 at 22:42
  • $f=1$ 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 $1$. – Mark Viola Jul 14 '15 at 22:52
  • Yes the local maximum is at $x = (1 + \sqrt{2/n})/(n-2)$ and the value there is less than $1$ if $n \geqslant 4$. – RRL Jul 14 '15 at 22:55
  • Is showing $f_n≤1$ for large n enough to satisfy the conditions for using the Dominated Convgerence Theorem? It does not need to be bounded for all n, not just large n? / For #2, is the change of variables necessary? We already have that $|f_{n}(x)|=|ne^{-nx}\sin(1/x)| \leq ne^{-nx} \leq \frac{1}{x}$, and so dominated by $g(x)=\frac{1}{x}$. Hence, dominated convergence applies. Am I missing something that makes the substitution necessary or is it just a cleaner bound? – Math244 Jul 14 '15 at 23:34
  • Never mind, I see why. $\int_{0}^{\infty} \frac{1}{x}$ does not converge. – Math244 Jul 14 '15 at 23:38
  • @Dr.MV: I don't believe $f=1$ is a global max for all $n \geq 1$. When $n=2$ the maximum on $(0,1]$ is 1.25 at $x=1$. – Math244 Jul 15 '15 at 12:50