Question

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

Answer 1

are you allowed to use lhopital?

Answer 2

everybody jumps for l'hospital... that should really be a last resort

Answer 3

is this\[\lim_{x\to0}{1-\cos^6x\over x\sin^5x}\]?

Answer 4

I think it would make this easier to use l'hopitals rule for this on an exam though, just to save time.

Answer 5

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.

Answer 6

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?

Answer 7

is it the first way I wrote it or the second?
are 5 and 6 exponents, or coefficients of the argument?

Answer 8

it's the second way

Answer 9

first multiply by\[\frac{1+\cos(6x)}{1+\cos(6x)}\]

Answer 10

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.

Answer 11

okay got it now, thank you so much ^^

Answer 12

welcome :D

Answer 13

alternative :
use the identity :
sin^2 (3x) = (1-cos(6x))/2 ---> 1 - cos(6x) = 2 sin^2 (3x)
so, it can be :