I need to compute the expression $xe^{-x/a}$ with the limits of $x$ from infinity to $0$. When I use L'Hopital's Rule, however, I do not get the correct answer of $0$.
What is the problem?
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First, write your expression as $x\over e^{x/a}$ (I'm assuming you are computing $\lim\limits_{x\rightarrow\infty} xe^{-x/a}$). – David Mitra Sep 21 '13 at 15:28
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Actually, I'm trying to evaluate the expression at x=infinity and subtract from it the expression evaluated at x=0. – Sep 21 '13 at 15:30
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A word for the future: We can prove that for positive reals $a$ and $b$ $$\lim_{x\longrightarrow \infty}{\frac{x^a}{e^{bx}}=0}$$ – hrkrshnn Sep 21 '13 at 15:48
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I think you mean: $$xe^{-x/a}|_{+\infty}^0=-xe^{-x/a}|_0^{+\infty}$$ If so, then we have $$-xe^{-x/a}|_0^{+\infty}=-0\times e^0+\frac{t}{e^{t/a}}|_{t\to +\infty}=\frac{a}{e^{t/a}}|_{t\to +\infty}=0$$
Mikasa
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