Question about calculation of the limit Indeterminate

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quangngaipro
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Question about calculation of the limit Indeterminate

Postby quangngaipro » Tue Jul 07, 2015 8:49 am UTC

Hello everybody.

Given the question
limx→0+xsin(x)
I have deducted so far that this has the indeterminate form 00 so I have taken the natural logarithm of both sides to give me:
limx→0+ln(y)=limx→0+sin(x)ln(x)
This still has an indeterminate form so I rearrange it to start using L'hospitals rule:
limx→0+ln(x)csc(x)
=limx→0+1x−csc(x)cot(x)
=limx→0+−sin(x)tan(x)x
=limx→0+−cos(x)sec2(x)1=−1
limx→0+xsin(x)=1e
I don't think this is correct, am I doing something wrong, and if so, where?
Last edited by quangngaipro on Thu Oct 15, 2015 7:47 am UTC, edited 2 times in total.

skullturf
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Re: Question about calculation of the limit Indeterminate

Postby skullturf » Wed Jul 08, 2015 4:37 pm UTC

0 divided by 0 is a famous indeterminate form, but 0 times 0 is not indeterminate.

If you have x times sin(x), that's of the form f(x) times g(x), where both f and g are approaching 0 as x approaches 0. Hence the product x*sin(x) also approaches 0.

Cauchy
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Re: Question about calculation of the limit Indeterminate

Postby Cauchy » Wed Jul 08, 2015 7:55 pm UTC

He's actually doing x^sin(x), but none of his binary operators rendered correctly in wherever he copied this from, I'm guessing.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

Nitrodon
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Re: Question about calculation of the limit Indeterminate

Postby Nitrodon » Wed Jul 08, 2015 8:03 pm UTC

With that in mind, it looks like your mistake is in the second-last line, where you differentiate -sin(x)tan(x) to obtain -cos(x)sec2(x). This derivative should be computed using the product rule instead.


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