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I'm trying to find this limit $$\lim \limits_{x \to \infty} x\int_{0}^{x}e^{t^2-x^2}dt$$ From the graph I can see that it equals $1/2.$

I've looked into making substitution in order to modify the integral to apply l'Hopital rule. I've tried $tx^2=s$ in order to move $x$ to the denominator but I wasn't able to get anything sensible.

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1 Answer 1

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Use l'Hospital and fundamental theorem of calculus:

$$\begin{align}\lim_{x \to \infty} x\int_{0}^{x}e^{t^2-x^2}\mathrm dt &=\lim_{x \to \infty} \frac{\int_{0}^{x}e^{t^2}dt}{e^{x^2}/x} =\lim_{x \to \infty} \frac{e^{x^2}}{\frac{(2x^2-1)e^{x^2}}{x^2}}\\ &=\lim_{x\to\infty}\frac{x^2}{2x^2-1}=\frac12. \end{align}$$

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