2

$\lim_{x\to 0^+} \frac{1}{x}=\infty$

But with l'Hospital's Rule

$\lim_{x\to 0^+} \frac{1}{x}=\lim_{x\to 0^+} \frac{0}{1}=0$

So where's my naive mistake?

Korgan Rivera
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2 Answers2

8

I believe l'Hopital's rule works only for indeterminate forms.

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    More precisely, $\lim_{x \to 0+} \dfrac{f(x)}{g(x)} = \lim_{x \to 0+} \dfrac{f'(x)}{g'(x)}$ if

    1. the limit on the right exists or is $+\infty$ or $-\infty$, and

    2. Either $\lim_{x \to 0+} f(x) = \lim_{x \to 0+} g(x) = 0$ or $\lim_{x \to 0+} g(x) = \pm \infty$.

     – Robert Israel Aug 23 '12 at 23:50
  • Ah, of course. How silly of me. :) Thanks! – Korgan Rivera Aug 23 '12 at 23:55
4

l'Hospital's Rule works on limits of the form "$\frac{0}{0}$" and "$\frac{\infty}{\infty}$" so you can not use it to evaluate your limit.

You can find the complete list of conditions you need to verify befote using l'Hospital's Rule here

Belgi
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  • Those conditions are not really complete. The rule works if the limit of $g$ is $+\infty$ or $-\infty$ no matter what the limiting behaviour of $f$ is. – Robert Israel Aug 24 '12 at 00:00
  • @RobertIsrael Wouldn't the above question be a counterexample to your claim? (Or, more likely, did I misunderstand what you meant?) – mboratko Aug 24 '12 at 00:48
  • @MichaelBoratko: this is talking about a limit of $f/g$. In the case at hand, $f = 1$ and $g = x$, and the limit of $g$ as $x \to 0$ is $0$, not $\infty$. Or you could take $f = 1/x$ and $g = 1$, and the limit of $g$ as $x \to 0$ is $1$. – Robert Israel Aug 24 '12 at 04:57
  • If you want to take $x \to \pm\infty$ with $f=1$ and $g=x$, the limit of $g$ is $\pm\infty$, so the rule should apply, and does: $$\lim_{x \to \pm \infty} \frac{1}{x} = \lim_{x \to \pm \infty} \frac{0}{1} = 0$$ – Robert Israel Aug 24 '12 at 04:59
  • @RobertIsrael Yes, my mistake. Thanks! – mboratko Aug 24 '12 at 12:03