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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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    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 $0$.2012-03-23

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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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    @Hunter For a statement of the more general L'Hospital rule (from Rudin's PoMA) and some examples [see my answers/comments here.](http://www.google.com/search?q=site:math.stackexchange.com+dubuque+rudin+hospital)2012-03-25