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I came across the following problem in my self-study, and wanted to know how to use Lebesgue Dominated Convergence to compute any of the following limits:

(a) $\lim\limits_{n \rightarrow \infty}$ $\int_0^\infty$ $(1+(x/n))^{-n} \sin (x/n)dx$

(b) $\lim\limits_{n \rightarrow \infty}$ $\int_0^1$ $(1+nx^{2})(1+x^2)^{-n}dx$

(c) $\lim\limits_{n \rightarrow \infty}$ $\int_0^\infty$ $n \sin (x/n) [x(1+x^2)]^{-1}dx$

Any help is greatly appreciated.

Update: I think I have successfully worked out arguments for each of (a) and (b), so I am less concerned about answers/strategies to those parts. However, (c) seems more tricky than the others, so if anyone visiting today sees how to handle (c) (in particular, a nice-enough dominating function!), let me know as it would be greatly appreciated.

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up vote 3 down vote accepted

Hint to (c):

  • What can we say about the function $$g(t)=\frac{\sin t}{t}\qquad\text{?}$$

  • What can we say about the function $$h(x)=\frac{1}{1+x^2}\qquad\text{?}$$

Edit: Adding an extra hint... $$n\sin(x/n)[x(1+x^2)]^{-1}=g(x/n)h(x).$$

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I'm assuming you are using Folland in your study.

(a) Let $f_n(x)= (1+(x/n))^{-n}\sin(x/n)$ and $g_n(x)= (1+(x/n))^{-n}$. Observe that $\lim_{n\to\infty}(1+(x/n))^{-n}=e^{-x}$ and $\lim_{n\to\infty} f_n(x)=0$. Since $|f_n(x)|\leq g_n(x)$ and it is easy to see that $\int g_n\to 1$, then by LDCT (the version of exercise 21 chapter 2) it follows that $\int f_n\to 0$.

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Since the sequence is decreasing, your inequality won't be true. – Robert Israel Feb 12 '12 at 6:16
You right, It is not as straightforward as I though. I'll edit with more details. – azarel Feb 12 '12 at 6:45
You can use $(1+x/2)^{-2}$ as your bound. – Robert Israel Feb 12 '12 at 6:50
@azarel How can you apply dominated convergence theorem on $g_n(x)$ to show that $\int g_n(x) \to 1$? It is not true that: $$\dfrac{1}{\left(1+\frac{x}{n}\right)^n} \le \dfrac{1}{e^x}$$ hence which function do you take to bound $g_n(x)?$ – Bman72 Jun 5 '14 at 12:54

For part (c), you can use the following inequality (which is not hard to verify) $$ |\sin(x)| \leq x\quad \forall x\in[0,\infty) $$ Then, we can get a dominating function as follows $$ \left|\frac{n\sin(x/n)}{x(1+x^2)}\right| \leq \frac{n(x/n)}{x(1+x^2)} = \frac{1}{1+x^2} $$

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