$\lim_{n\to\infty}\int_0^{\infty}\frac{n\sin y}{ny(1+n^2y^2)}n\,dy$ via DCT?

Question

I'm looking to calculate these limits/integrals:

$$\lim_{n\to\infty}\int_0^{\infty}\dfrac{n\sin (x/n)}{x(1+x^2)}\,dx\tag{1}$$

$$\lim_{n\to\infty}\int_0^{\infty}\dfrac{\sin(x/n)}{(1+x/n)^n}\,dx\tag{2}$$

I posted both since they look to be similar in nature and I am hoping if I can get a hint for either one of them ,I will be able to solve the other one.

So far what I have tried is to make a substitution of variables with $y=x/n$ which transforms both of the above into:

$$\lim_{n\to\infty}\int_0^{\infty}\frac{n\sin y}{ny(1+n^2y^2)}n\,dy$$

and $$\lim_{n\to\infty}\int_0^{\infty}n\dfrac{\sin y}{(1+y)^n}\,dy$$

I suspect Lebesgue's Dominating Convergence Thm should be used here but I am having difficulty selecting such a dominating $L_1$ function. Any help would be appreciated.

Answer 1

Hint for the first one:

Since $|\sin(\theta)| < \theta$ for $\theta>0$:

$$\left|\frac{n\sin(x/n)}{x(1+x^2)}\right| < \frac{n(x/n)}{x(1+x^2)} = \frac{1}{1+x^2}$$

For the second one:

Let $$f_n(x) = \frac{\sin(x/n)}{(1+x/n)^n}$$ We have that $\lim_{n\to\infty} f_n(x) = 0$. But $$|f_n(x)| \leq \frac{1}{(1+x/n)^n} \leq \frac{1}{(1+x/2)^2}$$ for $n\ge 2$.

So, for $n\ge 2$, $f_n(x)$ is dominated by $\frac{1}{(1+x/2)^2}$.

Answer 2

For the first part, set $f_n(x) := \dfrac{n\sin (x/n)}{x(1+x^2)}$. One has $f_n(x) \sim_{n\infty} \frac{1}{1+x^2}$. Now try to prove that $f_n$ is bounded (uniformly in $n$).