## anonymous 5 years ago LIMIT SIN^2(X)/X(1+COS(X) AS X GOES TO 0

1. anonymous

$\lim_{x\rightarrow 0}\frac{\sin^2x}{x(1+\cos x)}$ is it correct?

2. anonymous

YES

3. anonymous

zero

4. anonymous

5. anonymous

its not a indeterminate form , we can just sub in x=0

6. anonymous

Ok, answer is 0, but this is indeterminate form 0/0

7. anonymous

ahh yeh , im an idiot , didnt see the x factor on the bottom differnetiate top and bottom and try again

8. anonymous

YES I GOT 0/0 BUT MY TEACHER MARK MY PAPER STILL WRONG WITH MY STEPS I JUST SUBSTITUTE X FOR 0

9. anonymous

$=\lim_{x \rightarrow 0}\frac{\sin x ^{2}}{x} * \lim_{x \rightarrow 0}\frac{1}{1+\cos x} = \lim_{x \rightarrow 0}2\sin x \cos x * \frac{1}{2} = 0$

10. anonymous

so [ 2sinxcosx ] / [ x(-sin(x) ) + (1+cos(x)) ] 2sin(x)cos(x) / [ 1 +cos(x) -xsin(x) ]

11. anonymous

now when you sub x=0 , it isnt indeterminate , and we do get 0 as the final answer

12. anonymous

dumbcow, $\lim_{x\rightarrow a}f(x)g(x)\neq\lim_{x\rightarrow a}f(x)\cdot\lim_{x\rightarrow a}g(x)$

13. anonymous

^doesnt it? I thought I remember reading somewhere that it does

14. anonymous

really? are you sure lim x^2 as x->2 is 4 limx * limx as x->2 is 2*2=4

15. anonymous

no they can be separated, see below http://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx

16. anonymous

YES GUYS YOU CAN SEPARATE IM SURE OF THAT

17. anonymous

before you answer, shut the caps button off LOL!

18. anonymous

oh well ~