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I know the limit in the form infinite/infinite or 0/0 most of the time exist by l'hopital's rule, however if the limit is in the form of 0/infinite or infinite/0, is it undefined?

you could see l'hopital's rule here: http://en.wikipedia.org/wiki/L%27_Hopital%27s_Rule

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Very informally: $0/\infty$ is something small divided by something big. This will give you 0 as the limit. $\infty/0$ is trickier. One case: If the denominator is always positive, you have something big divided by something small and positive. What would this give you? –  David Mitra Mar 23 '12 at 20:20
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If the limit is $0/\infty$, then the limit exists and is equal to $0$ (if you have a quantity that is getting smaller and smaller, divided by a quantity that is getting larger and larger, then the quotient goes to $0$). If you have $\infty/0$, then the limit does not exist; it may be $\infty$, $-\infty$, or neither, depending on how the denominator approaches $0$. –  Arturo Magidin Mar 23 '12 at 20:21

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If as $x\to a$, $f(x)\to 0$ while $g(x)\to\infty$, then $\frac{f(x)}{g(x)}\to 0$. No L'Hospital's Rule needed, indeed L'Hospital's Rule could very well give the wrong answer, so must not be used.

The case "$\infty/0$" is more complicated. Again, L'Hospital's Rule is irrelevant, and must not be used. Suppose that as $x\to a$, $f(x)\to\infty$ and $g(x)\to 0$. If $g(x)$ approaches $0$ through positive values, one can conclude immediately that $\frac{f(x)}{g(x)}\to\infty$. If $g(x)$ approaches $0$ through negative values, one can similarly conclude that $\frac{f(x)}{g(x)}\to -\infty$. If, arbitrarily close to $x=a$, $g(x)$ can take on both positive and negative values, then the limit of $\frac{f(x)}{g(x)}$ does not exist.

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You can use L'Hopital's rule for limits of the form $a\over\infty$. All that's required is that the denominator tends to $\infty$ (though most texts I've seen don't point this out). –  David Mitra Mar 23 '12 at 20:39
    
@DavidMitra What would be the benefit of using L'Hoptial's rule if you can clearly see that the bottom function is growing faster? –  Hunter McMillen Mar 23 '12 at 21:19
    
@HunterMcMillen None, of course, if you have a $a/\infty$ form. (I think though, no hypothesis on the numerator is needed, other than being differentiable near the limit point. The more general rule might prove useful then.) I'm still wondering about the statement "L'Hopital's rule could very well give you the wrong answer"; which is why I raised the issue. –  David Mitra Mar 23 '12 at 21:24
    
@Hunter For a statement of the more general L'Hospital rule (from Rudin's PoMA) and some examples see my answers/comments here. –  Bill Dubuque Mar 25 '12 at 2:29

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