limits of 2 lebesgue integrals 
Compute the following limits, with justification, where the integrals denote Lebesgue integrals:


*

*$\lim\limits_{m\to\infty} \int_0^\infty \dfrac{m\sin (y/m)}{y(1+y^2)} dy$.

*$\lim\limits_{m\to\infty} \int_0^1 \dfrac{1+my^2}{(1+y^2)^m}dy.$
For 1), I think it's useful to use the Taylor expansion of $\sin(x) = x - \frac{x^3}{3!} + \dfrac{\sin c}{4!} x^4,$ where $c\in (0,x)$. Fix $m\ge 1$. Applying the above expansion to $\sin(y/m),$ we get $\frac{y}m - \frac{(y/m)^3}{3!} + \frac{(y/m)^4}{4!}\ge \sin(y/m) \ge \frac{y}m - \frac{y^3}{3! m^3} - \frac{(y/m)^4}{4!}$. Hence letting $g_m(y) = \frac{y}m - \frac{y^3}{3! m^3}, f_m(y) =  \dfrac{m\sin (y/m)}{y(1+y^2)}$, we have that for all $y\ge 0,$ $h_m^-(y) := \dfrac{1-y^2/(6m^2) - y^3/(24m^3)}{1+y^2}= \dfrac{m(g_m(y) - \frac{1}{24}(y/m)^4)}{y(1+y^2)} \leq f_m(y)\leq \dfrac{m(\frac{y}m - \frac{y^3}{3! m^3} + \frac{1}{24}(y/m)^4)}{y(1+y^2)} =: h_m^{-}(y).$
Note that both $h_m^-(y)$ and $h_m^+(y)$ converge pointwise to $\frac{1}{1+y^2}$ as $m\to \infty$, which implies by the Squeeze theorem that $\dfrac{m \sin(y/m)}{y(1+y^2)}$ converges pointwise to $\frac{1}{1+y^2}$ as $m\to\infty$. Also, $|\dfrac{m\sin(y/m)}{y(1+y^2)}|\leq |\dfrac{m(\frac{y}m - \frac{y^3}{3! m^3} + \frac{1}{24}(y/m)^4)}{y(1+y^2)}|$. But I need to upper bound $f_m(y)$ by an integrable function (a function $g$ so that $\int_0^\infty |g| dy < \infty$) to use the Dominated convergence theorem.
For the second limit, for any $y > 0, $ one has L'Hopital's rule that $\lim\limits_{m\to\infty} \dfrac{1+my^2}{(1+y^2)^m} = \lim\limits_{m\to\infty} \dfrac{my^2}{(1+y^2)^m} = 0.$ The issue is that to apply the Dominated convergence theorem, I need to find an upper bound for the integrand that's integrable.

For 1) use the fact that $|\sin t| \leq t$ for all $t >0$ so a dominating function is $\frac 1 {1+y^{2}}$.
For 2) use the fact that $(1+y^{2})^{m} \geq 1+my^{2}$ so a dominating function is the constant function $1$.
[The function $(1+x)^{m}-1-mx$ vanishes when $x=0$ and its dervative is $m[(1+x)^{m-1}-1]$ which is non-negative for $m >1, x \geq 0$. Hence, $(1+x)^{m}-1-mx\geq 0$].
