Consider the limit $\displaystyle\lim_{x\to\infty}\sqrt{x}e^{-x/2}$. Direct substitution gives an indeterminate product. By converting this indeterminate product into an indeterminate quotient, the problem is easily solved with l'Hopital's rule: $$\displaystyle\lim_{x\to\infty}\sqrt{x}e^{-x/2}=\displaystyle\lim_{x\to\infty}\frac{\sqrt{x}}{e^{x/2}}=\displaystyle\lim_{x\to\infty}\frac{1}{\sqrt{x}e^{x/2}}=0.$$

You could also obtain an indeterminate quotient using a different arrangement, specifically: $$\displaystyle\lim_{x\to\infty}\sqrt{x}e^{-x/2}=\lim_{x\to\infty}\frac{e^{-x/2}}{x^{-1/2}}.$$ However, l'Hopital's rule seems unable to resolve this limit until I rearrange things so that the exponential function is in the denominator.

Am I missing something here? I'm surprised that the second way doesn't work more easily.

  • 1
    $\begingroup$ General rule: If it doesn't look like its coming out nicely, you might want to consider flipping some fractions around. $\endgroup$ – Simply Beautiful Art Oct 16 '17 at 16:37
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    $\begingroup$ There are some simpler examples of L'Hospital's rule failing. Here's one:$$\lim_{x\to\infty}\frac{e^x}{e^x}$$ $\endgroup$ – Simply Beautiful Art Oct 16 '17 at 16:37

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