anonymous
  • anonymous
Find the limit lim {x->0} (1-cos6x)/(x sin5x) How to solve this?
Calculus1
chestercat
  • chestercat
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anonymous
  • anonymous
are you allowed to use lhopital?
TuringTest
  • TuringTest
everybody jumps for l'hospital... that should really be a last resort
TuringTest
  • TuringTest
is this\[\lim_{x\to0}{1-\cos^6x\over x\sin^5x}\]?

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abb0t
  • abb0t
I think it would make this easier to use l'hopitals rule for this on an exam though, just to save time.
TuringTest
  • TuringTest
Often you are not allowed to use l'hospital if the problem is\[\lim_{x\to0}{1-\cos(6x)\over x\sin(5x)}\]it can be solved fairly easily without l'hospital. so far, however, @tamiashi has not replied to me, so I don't want to go ahead with either until I get a response.
anonymous
  • anonymous
Sorry for being late in response, I'm not allowed to use L'hospital Can you show me how to solve it without L'hospital?
TuringTest
  • TuringTest
is it the first way I wrote it or the second? are 5 and 6 exponents, or coefficients of the argument?
anonymous
  • anonymous
it's the second way
TuringTest
  • TuringTest
first multiply by\[\frac{1+\cos(6x)}{1+\cos(6x)}\]
TuringTest
  • TuringTest
you then have\[{1-\cos^2(6x)\over x\sin(5x)(1+\cos(6x))}=\frac{1-\cos^2(6x)}{x}\cdot\frac1{\sin(5x)}\cdot\frac1{1+\cos(6x)}\]now multiply by \(\frac xx\) to get\[{1-\cos^2(6x)\over x\sin(5x)(1+\cos(6x))}=\frac{1-\cos^2(6x)}{x^2}\cdot\frac x{\sin(5x)}\cdot\frac1{1+\cos(6x)}\] manipluate this through trig identities and algebra sp that you can utilize\[\lim_{x\to0}\frac{\sin x}x=1\]to get your answer.
anonymous
  • anonymous
okay got it now, thank you so much ^^
TuringTest
  • TuringTest
welcome :D
RadEn
  • RadEn
alternative : use the identity : sin^2 (3x) = (1-cos(6x))/2 ---> 1 - cos(6x) = 2 sin^2 (3x) so, it can be : |dw:1361606909900:dw|

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