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pop000 · Jan 17, 2011, 05:55 AM

I have an Function (in the picture) and I need to Prove if X=0 can be a Vertical asymptotic
so I tried to use in L'HÃ´pital's rule but is not Possible Because I not get 0/0 or -+Infinity/+-Infinity.

so how can I prove it?

and more little thing tell me if I correct that X=1 is can't be an Vertical asymptotic

thanks for help :)

jcaron2 · Jan 17, 2011, 07:12 AM

If you rewrite the cot function as cos/sin, you'll see that the denominator really DOES go to zero.

pop000 · Jan 17, 2011, 07:56 AM

so there is not Vertical asymptotic for x=0 what about x=1 I calculated it and I found that there is no Vertical asymptotic when x=1 is correct?
thank you

jcaron2 · Jan 17, 2011, 08:46 AM

Yes, correct in both cases. Good job! But there ARE asymptotes at x= +/- 2, +/- 4, +/- 6,.

pop000 · Jan 17, 2011, 10:31 AM

hi. I still dosen't Success to calculated for x=0 can you Please show me any way to prove that there is no Vertical asymptotic for x=1 ?

really thank you.

pop000 · Jan 17, 2011, 10:41 AM

oh sorry I mean if you can show me any way to prove that there is no Vertical asymptotic for x=o.

thanks.

ebaines · Jan 17, 2011, 10:52 AM

pop000: Use l'Hospital's rule. The derivative of the numerator for x=0 is 1, and the derivative of the denominator for x = 0 is -pi/2. So the limit as x goes to 0 is -2/pi.

Unknown008 · Jan 17, 2011, 11:07 AM

That image is small >.<

${f(x) = \frac{x \cot $\frac{\pi x}{2}$}{x^3 - 1}}$

A strange graph that is... (http://www.wolframalpha.com/input/?i=y+%3D+\frac{x+cot+(\frac{\pi+x}{2})}{x^3+-+1})

Applying L'Hopital's rule (if I do that right... )

${f(x) = \frac{x \cot $\frac{\pi x}{2}$}{x^3 - 1} \rightarrow \frac{-1}{3x \sin$\frac{\pi x}{2}$}}$

But that graph is another strange one :eek:

ebaines · Jan 17, 2011, 11:22 AM

Here's how I applied l'hospital:

$ \lim_{x \to 0} \ \frac {x \cos(\frac {\pi x } 2) } {\sin(\frac {\pi x } 2) (x^3 -1)} \ = \\ \lim_{x \to 0} \ \frac {x (-sin(\frac {\pi x} 2) + \cos( \frac {\pi x} 2)} {\sin(\frac {\pi x} 2) (3x^2) + \frac {\pi} 2 \cos(\frac {\pi x} 2) (x^3-1)} \ = \ \frac 1 {-\pi/2} \ =\ \frac {-2} {\pi} $

Unknown008 · Jan 17, 2011, 11:28 AM

Ah, I kept the cot in the numerator and gave me -1/sin^2(pi x/2)

It seems strange to me though of how a different result is obtained beginning with the same function but applying l'Hopital's rule a 'different' way...

ebaines · Jan 17, 2011, 12:07 PM

Applying l'Hospital to the equation in its original form doesn't work too well:

$ \lim_{x \to 0} \ \frac {x \cot(\frac {\pi x } 2) } {(x^3 -1)} \ = \\ \lim_{x \to 0} \ \frac {x \frac {\pi} 2 \frac {-1} {sin ^2(\frac {\pi x} 2) }+ \cot( \frac {\pi x} 2)} {3x^2} \ = \ \frac {-\infty + \infty} 0 $

You can apply l'Hospital again, but it gets really messy and I gave up on it.

Unknown008 · Jan 17, 2011, 12:09 PM

Oops, seems I didn't do it well :o

Thanks!

pop000 · Jan 17, 2011, 12:22 PM

Yes I also saw this graph and I am not see there any Vertical asymptotic. :)

pop000 · Jan 17, 2011, 12:40 PM

Nice, thank.