Prove the limit problems I got two problems asking for the proof of the limit: 

Prove the following limit: $$\sup_{x\ge 0}\ x e^{x^2}\int_x^\infty e^{-t^2} \, dt={1\over 2}.$$

and, 

Prove the following limit: $$\sup_{x\gt 0}\ x\int_0^\infty {e^{-px}\over {p+1}} \, dp=1.$$

I may feel that these two problems are of the same kind. World anyone please help me with one of them and I may figure out the other one? Many thanks! 

After a suitable substitution, both limits are easily handled by Dominated Convergence or Monotone Convergence.

The First Integral
Substituting $t\mapsto\sqrt{t+x^2}$,
$$
\begin{align}
xe^{x^2}\int_x^\infty e^{-t^2}\,\mathrm{d}t
&=xe^{x^2}\int_0^\infty e^{-t-x^2}\frac{\mathrm{d}t}{2\sqrt{t+x^2}}\\
&=\frac12\int_0^\infty e^{-t}\frac{\mathrm{d}t}{\sqrt{1+t/x^2}}\tag{1}
\end{align}
$$
Equation $(1)$ shows that the integrand is increasing with $x$ and by Dominated or Monotone Convergence, the limit is
$$
\frac12\int_0^\infty e^{-t}\,\mathrm{d}t=\frac12\tag{2}
$$

The Second Integral
Substituting $p\mapsto p/x$,
$$
\begin{align}
x\int_0^\infty\frac{e^{-px}}{p+1}\,\mathrm{d}p
&=\int_0^\infty\frac{e^{-p}}{1+p/x}\,\mathrm{d}p\tag{3}
\end{align}
$$
Equation $(3)$ shows that the integrand is increasing with $x$ and by Dominated or Monotone Convergence, the limit is
$$
\int_0^\infty e^{-p}\,\mathrm{d}p=1\tag{4}
$$
