Find without L'Hopital : $\lim_{x\rightarrow\infty}xe^{x^2} \left(∫_0^xe^{-t^2}dt-∫_0^∞e^{-t^2 } dt\right) = \ ?$ $$\lim_{x\rightarrow\infty}xe^{x^2} \left(∫_0^xe^{-t^2}dt-∫_0^∞e^{-t^2 } dt\right) = \ ?$$
I found this limit using L'Hopital's rule and error function. That is relatively easy I am wondering if there is any suggestion to find it without using L'Hopital's rule. 
The final answer is $-\frac{1}{2}$.  Another version:$$\lim_{x\rightarrow\infty}x2^{x^2} \left(∫_0^x2^{-t^2}dt-∫_0^∞2^{-t^2 } dt\right) = \ ?$$Suggested by Jalil Hajimir 
 A: Since you seem to be familiar with the error function, you face
$$y=xe^{x^2} \left(∫_0^xe^{-t^2}dt-∫_0^∞e^{-t^2 } dt\right)=\frac{\sqrt{\pi }}{2}x e^{x^2} \left( \text{erf}(x)-1\right)=-\frac{\sqrt{\pi }}{2}x e^{x^2}\text{erfc}(x)$$
Now, using the asymptotic series (see here)
$$\text{erfc}(x)\sim\frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2x^2)^n}$$ makes
$$y= -\frac{1}{2}+\frac{1}{4 x^2}-\frac{3}{8x^4}+O\left(\frac{1}{x^6}\right)$$ which shows the limit and how it is approached.
It also gives a good approximation value of $y$
A: Using integration by parts,
$$ \int_x^\infty e^{-t^2}\,\mathrm{d}t = \int_x^\infty \frac{2te^{-t^2}}{2t} \,\mathrm{d}t = \frac{e^{-x^2}}{2x} - \int_x^\infty \frac{e^{-t^2}}{2t^2} \,\mathrm{d}t  ,$$ and
$$ \int_x^\infty \frac{e^{-t^2}}{2t^2}\,\mathrm{d}t =  \frac{e^{-x^2}}{4x^3} - \int_x^\infty \frac{3e^{-t^2}}{4t^4}\,\mathrm{d}t. $$
Since the integrands are always non-negative, this gives $$\frac{e^{-x^2}}{2x} - \frac{e^{-x^2}}{4x^3}   \le \int_x^\infty e^{-t^2}\,\mathrm{d}t \le \frac{e^{-x^2}}{2x}.$$
Thus, $$ xe^{x^2}\left( \int_0^x e^{-t^2}\,\mathrm{d}t - \int_0^\infty e^{-t^2} \,\mathrm{d}t \right) = - xe^{x^2} \int_x^\infty e^{-t^2} \,\mathrm{d}t = -\frac12 + O(x^{-2}),$$ and the limit is $-1/2$. 
It is straightforward to replace the $e$s in the above by any constant $a > 1$, since $a^{x^2} = e^{ (\sqrt{\ln(a)} x)^2}$, at which point you can rescale $y = \sqrt{\ln(a)} x$ and proceed as above. 
