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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 $-\infty$ Graph

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$$\lim_{x\rightarrow \infty}\large \frac{\int^{x}_{0}e^{t^3-x^3}dt}{x^{-2}} = \lim_{x\rightarrow \infty}\frac{\int^{x}_{0}e^{t^3}dt}{x^{-2}e^{x^3}}$$

Using L'hopital's rule

$$\large \lim_{x\rightarrow \infty}\frac{e^{x^3}}{x^{-2}\cdot e^{x^3}\cdot 3x^2-e^{x^3}\cdot 2x^{-3}} = \frac{1}{3}$$

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  • $\begingroup$ In the 2nd step of 1st line ,how did you bring $e^{x^3}$ to denominator? $\endgroup$ – MatheMagic Jan 28 '17 at 5:18
  • $\begingroup$ It's not dependant on the variable of integration. $\endgroup$ – Plopperzz Jan 28 '17 at 5:21
  • $\begingroup$ 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. $\endgroup$ – dark32 Jan 30 '17 at 11:33
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With L'Hospital's rule \begin{eqnarray} \ell&=& \lim_{x\to \infty} x^2\int_{0}^{x} e^{t^3- x^3}dt\\ &=& \lim_{x\to \infty} \frac{x^2\int_{0}^{x} e^{t^3}dt}{e^{x^3}}\\ &=& \lim_{x\to \infty} \frac{2x\int_{0}^{x} e^{t^3}dt+x^2e^{x^3}}{3x^2e^{x^3}}\\ &=& \lim_{x\to \infty} \frac{2x\int_{0}^{x} e^{t^3}dt}{3x^2e^{x^3}}+\lim_{x\to \infty}\frac{x^2e^{x^3}}{3x^2e^{x^3}}\\ &=& \frac23\ell\lim_{x\to \infty}\frac{1}{x^4}+\frac13\\ &=& \color{blue}{\frac13} \end{eqnarray}

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