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! 

For the first problem, use integration by parts, followed by crude estimation. 
Let $u=\frac{1}{t}$ and $dv=te^{-t^2}\,dt$. Then $du=-\frac{1}{t^2}\,dt$ and we can take $v=-\frac{1}{2}e^{-t^2}$. Thus our integral is equal to
$$\frac{1}{x}\cdot \frac{1}{2}e^{-x^2}-\int_x^\infty \frac{1}{2t^2}e^{-t^2}\,dt.$$
Multiplying $\frac{1}{x}\cdot\frac{1}{2}e^{-x^2}$ by $xe^{x^2}$ gives us the main term. It remains to check that $\int_x^\infty \frac{1}{2t^2}e^{-t^2}\,dt$ is  small enough that multiplying by $xe^{x^2}$ gives a small result. 
We have 
$$\int_x^\infty \frac{1}{2t^2}e^{-t^2}\,dt \lt \frac{1}{2x^2}\int_x^\infty e^{-t^2}\,dt.$$ 
We can bound $\int_x^\infty e^{-t^2}\,dt$ by integrating from $x$ to $x+1$, and from $x+1$ to $\infty$. The integral from $x$ to $x+1$ is $\lt e^{-x^2}$. 
For $x+1$ to $\infty$, make the change of variable $t=s+1$. We want
$\int_{s=x}^\infty e^{-s^2-2s-1}\,ds$, which is less than $e^{-x^2}\int_x^\infty e^{-2s-1}\,ds$.  The remaining integral is bounded by a constant. 
