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Good day

I'm currently doing some math homework (don't worry I won't ask anyone to solve anything) and I don't think I'm understanding limits correctly. More precisely how the l'Hôpital rule works.

I know I can/should be able to apply it if it is either $\infty/\infty$ or $0/0$ but I was wondering does it have anything to do with $0/\infty$ or $\infty/0$?
Anything you think might help me better understand limits would be appreciated.

P.S. I'm sorry if this is a simple question, math is not my strongest point.

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  • $\begingroup$ No, you don't use l'Hôpital on anything you can't reduce to the indeterminate $0/0$ or $\infty/\infty$ cases. Usually other techniques will be much more transparent than l'Hôpital anyway. $\endgroup$ – J. M. is a poor mathematician Oct 1 '11 at 14:47
  • $\begingroup$ Can you name the other techniques? And can I assume that if I have a limit of a function f(x) that is 0/infinity that it is just 0? $\endgroup$ – Siemsen Oct 1 '11 at 15:01
  • $\begingroup$ The "other techniques" come out on a case-to-case basis... $\endgroup$ – J. M. is a poor mathematician Oct 1 '11 at 15:16
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I was wondering does it have anything to do with 0/infinity or infinity/0?

No, because $0/\infty$ and $\infty/0$ are not indeterminate cases. Symbolically one can write $0/\infty=0\times 0=0$ and $\infty/0=\infty\times \infty=\infty$ (without sign).

For instance, the function $f(x)=1/x\to 0$ (as $x\to \infty$) and the function $g(x)=x\to \infty$, (as $x\to \infty$). And we have, as $x\to \infty$, $$f(x)/g(x)=(1/x)/x=1/x^2\to 0$$ and $$g(x)/f(x)=x/(1/x)=x^2\to \infty.$$

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  • $\begingroup$ @J.M. Do we write "indeterminate case" (and not "undetermined case")? $\endgroup$ – Américo Tavares Oct 1 '11 at 15:18
  • $\begingroup$ Indeterminate is the English term of art I'm accustomed to. (Maybe somebody else uses "undetermined", but I haven't seen 'em.) $\endgroup$ – J. M. is a poor mathematician Oct 1 '11 at 15:19
  • $\begingroup$ @J.M. Thanks! (I have corrected it). $\endgroup$ – Américo Tavares Oct 1 '11 at 15:21

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