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

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?

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l'Hopital's rule doesn't apply here. –  Qiaochu Yuan Aug 23 '12 at 23:46
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2 Answers 2

up vote 8 down vote accepted

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
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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

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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 0:00
    
@RobertIsrael Wouldn't the above question be a counterexample to your claim? (Or, more likely, did I misunderstand what you meant?) –  process91 Aug 24 '12 at 0:48
    
@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 4:57
    
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 4:59
    
@RobertIsrael Yes, my mistake. Thanks! –  process91 Aug 24 '12 at 12:03
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