Does l'Hopital's rule apply to -infinity/infinity?

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It works for 0/0 or infinity/infinity

Does it work for -infinity/infinity? wiki suggests no.

Books give examples for lim xln(x) and does it and appplies this rule. HOwever it's really -infinity/infinity!
 
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Of course it works for -infinity/infinity! Thats because (if and only if the limit exists) \lim_{x\to a} - f(x) = - \lim_{x\to a} f(x). The second part is one of the cases you know it works.
 
Are you talking about l'Hopital, or l'Hospital's rule? If so, that is a weird spelling in the title!
 
cristo said:
Are you talking about l'Hopital, or l'Hospital's rule? If so, that is a weird spelling in the title!
LOL! Damn ... I've never seen it butchered that badly. Good laugh, thanks :-] no offense either :p
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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