L'Hopital's Rule for Solving Indeterminate Form (-∞)/∞

  • Thread starter Thread starter kasse
  • Start date Start date
kasse
Messages
383
Reaction score
1
Can one use L'Hopital's Rule on the indeterminate form (-∞)/∞ ?

And by the way, is there a way to write mathematical signs like ∞, the integral sign etc except google-ing, cutting and pasting?
 
Last edited:
Physics news on Phys.org
I think you can. It would probably be:

-\lim_{x\rightarrow c} \frac{f(x)}{g(x)} so that both functions approach \infty

you write those signs in tags as follows: \infty \int
 
Last edited:
Yes, it's a french name. But that doesn't really help me a lot :-p
 
OK, thanks!
 
I'm not sure, I think its only 0/0, as it comes from the Taylor expansion about the point that x approaches, ie.

\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f(a) + f'(x)(x-a) + ...}{g(a) + g'(x)(x-a) + ...} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}

With f(a)=g(a)=0

Something like that anyway.
 
No you can use L'Hopitals Rule for indeterminate forms, not just 0/0.
 
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...

Similar threads

Back
Top