Find $\lim\limits_{t\to 0^+} \int_0^{\infty} \frac{t f(x)}{t^2+x^2} dx$ The question is:
Let $f(x)$ be bounded and continuous on $[0,\infty)$. Let $\displaystyle F(t)=\int_0^{\infty} \frac{t f(x)}{t^2+x^2} dx$ for $t>0$.
Find $\displaystyle \lim_{t\to 0^+} F(t)$.

If I set $|f(x)|\leq M$, then I can obtain $|F(t)|\leq \frac{\pi}{2} M$. But I can not find the limit.
I try to rewrite $\displaystyle F(t)=\int_0^{\infty} \frac{f(t y)}{1+y^2} dy$.
I think if I can put the limit into the integration, then $$\displaystyle \lim_{t\to o^+} F(t)=\int_0^{\infty} \frac{\lim\limits_{t\to 0^+}f(t y)}{1+y^2} dy=\frac{\pi}{2} f(0).$$
But I don't know whether I can do. I hope I can receive some hints or method in here.
Thanks for your attention.
 A: Split integral into two parts. Let $\epsilon > 0$ and let $\delta > 0$ such that $|x| \leq \delta$ implies $|f(x) - f(0)| \leq \epsilon$. Then
$$ \begin{align*}
\left| F(t) - \frac{\pi}{2} f(0) \right|
& = \left| \int_{0}^{\infty} \frac{f(t x) - f(0)}{1+x^2} \; dx \right| \\
& \leq \int_{0}^{\infty} \frac{|f(t x) - f(0)|}{1+x^2} \; dx \\
& = \int_{0}^{\delta / t} \frac{|f(t x) - f(0)|}{1+x^2} \; dx + \int_{\delta / t}^{\infty} \frac{|f(t x) - f(0)|}{1+x^2} \; dx \\
& \leq \epsilon \int_{0}^{\delta / t} \frac{dx}{1+x^2} + \int_{\delta / t}^{\infty} \frac{2M}{1+x^2} \; dx \\
& \leq \frac{\pi}{2} \epsilon + \int_{\delta / t}^{\infty} \frac{2M}{1+x^2} \; dx.
\end{align*}$$
Thus taking $\limsup_{t \to 0^+}$, we obtain
$$\limsup_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| \leq \frac{\pi}{2}\epsilon.$$
Since this is true for any $\epsilon > 0$, we must have
$$\begin{align*}
\limsup_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| = 0
& \Longrightarrow \lim_{t \to 0^+} \left| F(t) - \frac{\pi}{2} f(0) \right| = 0 \\
& \Longrightarrow \lim_{t \to 0^+} F(t) = \frac{\pi}{2} f(0)
\end{align*}$$
A: As you have done, by using the substitution $x=ty$, we can rewrite $F(t)$ as 
$\displaystyle F(t)=\int_0^{\infty} \frac{f(t y)}{1+y^2} dy$. By assumption, $f$ is bounded, i.e. $|f(ty)|\leq M$ for some constant $M$. This implies that 
$$\left|\frac{f(t y)}{1+y^2}\right|\leq \frac{M}{1+y^2}\mbox{ for all }y\in[0,\infty).$$
Note that the function $\displaystyle\frac{M}{1+y^2}$ is integrable on $[0,\infty)$, for 
$$\int_0^\infty \frac{M}{1+y^2}dy=\frac{M\pi}{2}<\infty.$$
By Dominated convergence theorem, we have
$$\lim_{t\rightarrow 0^+}F(t)=\lim_{t\rightarrow 0^+}\int_0^{\infty} \frac{f(t y)}{1+y^2} dy=
\int_0^{\infty} \lim_{t\rightarrow 0^+}\frac{f(t y)}{1+y^2} dy=
\int_0^{\infty} \frac{f(0)}{1+y^2} dy=\frac{f(0)\pi}{2}.$$
