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Find the limit lim {x->0} (1-cos6x)/(x sin5x) How to solve this?

Calculus1
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are you allowed to use lhopital?
everybody jumps for l'hospital... that should really be a last resort
is this\[\lim_{x\to0}{1-\cos^6x\over x\sin^5x}\]?

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Other answers:

I think it would make this easier to use l'hopitals rule for this on an exam though, just to save time.
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.
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?
is it the first way I wrote it or the second? are 5 and 6 exponents, or coefficients of the argument?
it's the second way
first multiply by\[\frac{1+\cos(6x)}{1+\cos(6x)}\]
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.
okay got it now, thank you so much ^^
welcome :D
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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