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'Hôpital'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 :)

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

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

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

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.

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

thanks.

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.

That image is small >.<



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... )



But that graph is another strange one :eek:

Here's how I applied l'hospital:

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...

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



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

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

Thanks!

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

Nice, thank.