# Evalutate $\lim_{x\to \infty} x^2 \int_{0}^{x} e^{t^3 - x^3}dt$

I need to evaluate the following limit: $$\lim_{x\to \infty} x^2 \int_{0}^{x} e^{t^3 - x^3}dt$$

I've tried L'hopital's rule but the answer comes out to be $-\infty$ $$\lim_{x\to \infty} \frac{\frac{d}{dx}{\int_{0}^{x}{e^{x^3 - x^3}}}{x}}{\frac{d}{dx}\left(\frac{1}{x^2}\right)}$$ $$\lim_{x\to \infty} \frac{3x^2e^{x^3 - x^3}}{\frac{-2}{x^3}}$$ $$\lim_{x\to \infty} -\frac{3x^5}{2} = -\infty$$

Graph of the function shows limit to be not $-\infty$

$$\lim_{x\rightarrow \infty}\large \frac{\int^{x}_{0}e^{t^3-x^3}dt}{x^{-2}} = \lim_{x\rightarrow \infty}\frac{\int^{x}_{0}e^{t^3}dt}{x^{-2}e^{x^3}}$$
$$\large \lim_{x\rightarrow \infty}\frac{e^{x^3}}{x^{-2}\cdot e^{x^3}\cdot 3x^2-e^{x^3}\cdot 2x^{-3}} = \frac{1}{3}$$
• In the 2nd step of 1st line ,how did you bring $e^{x^3}$ to denominator? – MatheMagic Jan 28 '17 at 5:18
• Thank you for the answer, I wish to know where my method went wrong. As it seemed to me that the numerator must tend to zero for the limit to exist. So $\frac{\int{e^{t^3-x^3}}}{\frac{1}{x^2}}$ must be in $\frac{0}{0}$ form. – dark32 Jan 30 '17 at 11:33