What is the limit using L'hospital's rule?

[Avi1995]

Using L'hospitals rule, find the limit

I seem to stuck using L'hospital's rule ,I replaced cotx=1/tanx but the derivatives of even 4th order are not simplying things.

 

[Dick]

Before you even start with l'Hopital's rule you need to check you can write the limit in an indeterminant form 0/0 or infinity/infinity. Did you check that? And do you really mean lim x->inf??

 

[Avi1995]

Before you even start with l'Hopital's rule you need to check you can write the limit in an indeterminant form 0/0 or infinity/infinity. Did you check that?

Yes,
put cotx=(tanx)^-1, we have then
tan^2x-x^2
--------------
x^2tan^2x
Then we can use the rule.

But I tried wolfram alpha just now and found out that this limit does not exist.
http://www.wolframalpha.com/input/?i=limit+x%5E%28-2%29-%28cotx%29%5E2%2C++x-%3Einf

But instead x->0 exists.
http://www.wolframalpha.com/input/?i=limit+x%5E%28-2%29-%28cotx%29%5E2%2C++x-%3E0
This is the same answer given in textbook. Probably a printing error.
Sorry If I wasted your time.

 

[Dick]

Yes,
put cotx=(tanx)^-1, we have then
tan^2x-x^2
--------------
x^2tan^2x
Then we can use the rule.

But I tried wolfram alpha just now and found out that this limit does not exist.
http://www.wolframalpha.com/input/?i=limit+x%5E%28-2%29-%28cotx%29%5E2%2C++x-%3Einf

But instead x->0 exists.
http://www.wolframalpha.com/input/?i=limit+x%5E%28-2%29-%28cotx%29%5E2%2C++x-%3E0
This is the same answer given in textbook. Probably a printing error.
Sorry If I wasted your time.

Yes, that's the way to set it up, and I'm sure they mean lim x->0. Given that you should get the right result after four derivatives.