$\lim\limits_{x\to \infty} x^2 \int_0^x e^{t^3-x^3} dt$

Question

Plz solve this integral $$\lim_{x\to \infty} x^2 \int_0^x e^{t^2-x^2} dt = \frac{1}{2}$$

I tried solving this for like 3 days but couldn't even find the required answer even on online platforms like Wolfram. Those of you who might doubt the authenticity of the question, I am attaching the source -

http://kvpy.iisc.ernet.in/main/2016-questionpapers.htm ; Choose stream SB/SX ; Question no. 12, page no. 5

Will attach image after gaining required points.

Answer 1

Let f(x)=$\int_0^x \exp(x^2) dx$ and $g(x)=x^{-1} \exp(x^2)$. Apply L'Hospital's Rule to $\frac f g$.

Answer 2

Indeed the limit you ask about is infinity. We can show this with L'Hospital's rule.

Let $$F(x) = x^2e^{-x^2}\int_0^x e^{t^2} dt$$

You are interested in $\lim_{x \rightarrow \infty} F(x)$. Set $$f(x) = \int_0^x e^{t^2} dt$$ $$g(x) = x^{-2} e^{x^2}$$

Then $F(x) = \frac{f(x)}{g(x)}$, and we see that $$\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} g(x) = +\infty$$

This is just the sort of indeterminate limit form that L'hospitals rule deals with.
We calculate that $$g'(x) = 2(x^{-1} - x^{-3})e^{x^2}$$ and $$f'(x) = e^{x^2}$$ (See here if you need a reference for differentiation under the limit sign.)

Hence $$\frac{f'(x)}{g'(x)} = \frac{1}{2} \frac{x^3}{x^{2} - 1}$$

and by L'Hospital's rule we have $\lim_{x \rightarrow \infty} F(x) = \lim_{x \rightarrow \infty} \frac{1}{2} \frac{x^3}{x^{2} - 1} = \infty$

That said, if $g(x)$ were instead $x^{-1} e^{x^2}$, you can see how things would work out nicely and the limit would be $\frac{1}{2}$