Limits with definite integration

Question

$Evaluate:$ $$\lim_{x\rightarrow \infty} \frac{x \int_{0}^{x} e^{x^2} dx} {e^{x^2}}$$

My attempt

First I dealt with numerator with integral :

As $$\int_{b}^{a} f(x) dx= \int_{b}^{a} f(a+b-x)\,dx$$ To get it as $$\int_{0}^{x} 1\cdot dx =x$$

So the limit turned to be

$$\lim_{n\rightarrow \infty} \frac {x^2} {e^{x^2}}=0$$

Is it true?

Answer 1

Using L'Hospital's Rule yields

$$\begin{align} \lim_{x\to\infty}\frac{x\int_0^x e^{t^2}\,dt}{e^{x^2}}&=\lim_{x\to\infty}\frac{\int_0^x e^{t^2}\,dt+xe^{x^2}}{2xe^{x^2}}\\\\ &=\frac12+\lim_{x\to\infty}\frac{\int_0^x e^{t^2}\,dt}{2xe^{x^2}}\\\\ &=\frac12+\lim_{x\to\infty}\frac{e^{x^2}}{2e^{x^2}+4x^2e^{x^2}}\\\\ &=\frac12 \end{align}$$

Answer 2

By L'Hospital,

$$\frac{\displaystyle\int_0^xe^{t^2}dt}{\dfrac{e^{x^2}}x}\to\frac{e^{x^2}}{\dfrac{(2x^2-1)e^{x^2}}{x^2}}\to\frac{x^2}{2x^2-1}\to\frac12.$$