## tamiashi 2 years ago Find the limit lim {x->0} (1-cos6x)/(x sin5x) How to solve this?

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1. completeidiot

are you allowed to use lhopital?

2. TuringTest

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

3. TuringTest

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

4. abb0t

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

5. 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.

6. tamiashi

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?

7. TuringTest

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

8. tamiashi

it's the second way

9. TuringTest

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

10. 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.

11. tamiashi

okay got it now, thank you so much ^^

12. TuringTest

welcome :D