L'hopital's Rule

Shelled · 2015-06-18 22:15:44

I applied L'hopital's three times where i got:

where I thought the last was a determinant form and got the final value = 0, but that's not correct. Could someone show me where i went wrong?

zetafunc · 2015-06-18 22:46:10

You can't apply L'Hopital a third time, because

is not an indeterminant form. In fact, with that limit, approaching 0 from the right and left gives you positive and negative infinity -- i.e. the limit doesn't exist.

Shelled · 2015-06-18 23:38:52

Thank you
would you determine that its negative & positive infinity by looking at the graph it makes?

zetafunc · 2015-06-19 01:48:26

That's one way, but you could note what happens for negative and positive x. Since cos(-x) = cos(x), then the function is negative for negative x, and positive for positive x (for x sufficiently close to 0).

You know it'll approach infinity (positive or negative) since cos(x) is bounded and 1/x goes to infinity as x goes to 0.

Shelled · 2015-06-19 23:14:30

Alright, thank you

Math Is Fun Forum

#1 2015-06-18 22:15:44

L'hopital's Rule

#2 2015-06-18 22:46:10

Re: L'hopital's Rule

#3 2015-06-18 23:38:52

Re: L'hopital's Rule

#4 2015-06-19 01:48:26

Re: L'hopital's Rule

#5 2015-06-19 23:14:30

Re: L'hopital's Rule

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