Question

Limits,

Help me solve this limit: http://screencast.com/t/4ufvmkIx

Answer 1

What is x tending to?

Answer 2

Pi-

Answer 3

This seems a limit you can solve using L'Hopital rule, but you'll have to modify a little the expression.

Answer 4

Can you help me a little please?

Answer 5

on the first steps

Answer 6

Do you see the part,
\[x-\pi\]?
This tend to 0. And the other part,
\[\tan(x/2)\]tends to infinity.

Answer 7

Yes ik

Answer 8

You can modify the second term and put like
\[1/\cot(x/2)\]
Do you see why is better to do it?

Answer 9

No please explain me why its better to do it ?

Answer 10

The idea is getting a limit that can be solved using L'Hopital rule, so it must be infinity/infinity or 0/0. In this case, I choose the second.
\[\lim_{x\rightarrow\pi^-}(x-\pi)\tan(x/2)=\frac{(x-\pi)}{\cot(x/2)}\]But now, as I have said, is 0/0, and we can apply L'Hopital rule. Do you understand?

Answer 11

So you have to solve now,
\[\lim_{x\rightarrow \pi^-}\frac{x-\pi}{\cot(x/2)}\]
I don't say it before but you should remember that,
\[\cot \alpha=\frac{1}{\tan\alpha}\]

Answer 12

I found the derivative of http://screencast.com/t/WoZiKKcDgpPu

Bad move I guess ?

Answer 13

You have to find the derivative of both parts (numerator and denominator) independently.

Answer 14

random question: can I convert this http://screencast.com/t/Yhyb1a000 to this http://screencast.com/t/Yhyb1a000 once again so that I have this ?? http://screencast.com/t/5vt7LmP0

Answer 15

that was stupid , I just realised -_- never mind

Answer 16

Do you know how to continue?

Answer 17

what's the derivative of cot ?? csc^2 ?

Answer 18

I am trying to derivative both up and down

Answer 19

Yes, but with a minus sign. ;)

Answer 20

Yes, this is what you have to do.

Answer 21

When you find the derivative, tell me result if you want a check of it.

Answer 22

ok

Answer 23

can you help me find the derivative of the denominator ?

Answer 24

Yes, it should be like that,
\[(\cot x)'=\left(\frac{\cos x}{\sin x}\right)'=\frac{-\sin x\sin x-\cos x\cos x}{\sin^2x}=-\frac{1}{\sin^2x}\]
This is the form I think is best for this problem. You can express it also like -csc^2x.

Answer 25

I miss something, in this particular case,
Yes, it should be like that,
\[(\cot (x/2))'=\left(\frac{\cos (x/2)}{\sin (x/2)}\right)'=\frac{-(1/2)\sin (x/2)\sin (x/2)-(1/2)\cos (x/2)\cos (x/2)}{\sin^2(x/2)}=\\
=-\frac{1}{2\sin^2x}\]
This is the form I think is best for this problem. You can express it also like -csc^2x.

Answer 26

So the problem should look like,
\[\lim_{x\rightarrow \pi^-}\frac{x-\pi}{\cot(/2)}=\lim_{x\rightarrow \pi^-}\frac{1}{-\frac{1}{2\sin^2 (x/2)}}\]

Answer 27

Do you understand it unitl now?

Answer 28

There is a final step,
\[\lim_{x\rightarrow\pi^-}-2\sin^2(x/2)=-2\]

Answer 29

thanks man.

Answer 30

;)