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

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1. completeidiot Group Title

are you allowed to use lhopital?

2. TuringTest Group Title

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

3. TuringTest Group Title

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

4. abb0t Group Title

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

5. TuringTest Group Title

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 Group Title

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 Group Title

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

8. tamiashi Group Title

it's the second way

9. TuringTest Group Title

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

10. TuringTest Group Title

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 Group Title

okay got it now, thank you so much ^^

12. TuringTest Group Title

welcome :D