L'hospital's Rule Problem

  • Thread starter Avi1995
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  • #1
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Using L'hospitals rule, find the limit
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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.
 

Answers and Replies

  • #2
Dick
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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??
 
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  • #3
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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.
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
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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.
 

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