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

Shelled
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Registered: 2014-04-15
Posts: 44

L'hopital's Rule

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? smile

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

zetafunc
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Registered: 2014-05-21
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Re: L'hopital's Rule

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.

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

Shelled
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Registered: 2014-04-15
Posts: 44

Re: L'hopital's Rule

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

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

zetafunc
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Registered: 2014-05-21
Posts: 2,432
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Re: L'hopital's Rule

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.

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

Shelled
Member
Registered: 2014-04-15
Posts: 44

Re: L'hopital's Rule

Alright, thank you smile

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