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**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.**

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- Thread starter Avi1995
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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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Yes,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?

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.

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Dick

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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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