May de l'Hopital's rule be used (for $\frac{f(x)}{g(x)}$) if $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = -\infty$ (or vice versa)? Wikipedia seems to be quite ambiguous as it says $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = \pm \infty $.
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2Short answer: yes. (Longer answer: there's some messiness in your hypothesis. If someone is keen, they can write it out more carefully.) – Simon S Jun 08 '15 at 19:15
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yes sorry i mean lim f(x) etc. edited it – user246749 Jun 08 '15 at 19:16
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yes, but why or why? – abel Jun 08 '15 at 19:18
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@abel what do you mean – user246749 Jun 08 '15 at 19:23
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never had to use l'hopitals. – abel Jun 08 '15 at 19:24
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@abel Congratulations on using only spanners and screwdrivers. Most of us don't mind using spanners, screwdrivers and hammers. – Simon S Jun 08 '15 at 19:27
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Note that if $\lim_{x \to a} f(x) = \infty$ then $\lim_{x \to a} -f(x) = -\infty$. Thus if you can prove that $$\lim_{x \to a} \frac{-f'(x)}{g'(x)} \text{ exists, and } g'(x)\not = 0 \text{ for $x$ in a neighbourhood around $a$ except possibly in $a$.}$$ L'Hopital's rule gives us that $$ \lim_{x \to a} \frac{-f(x)}{g(x)} =\lim_{x \to a} \frac{-f'(x)}{g'(x)} = -\lim_{x \to a} \frac{f'(x)}{g'(x)} $$ Multiplying with -1 on both sides yields that $$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$
John
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