Question

evaluate limit as x approaches 0 sin^2x/(1-cosx)

Answer 1

If you use sin²x+cos²x=1 to replace sin²x with 1-cos²x, you can factor it and simplify.
After this, the limit is easy.

Answer 2

To factor 1-cos²x, remember it is like a²-b², and that can be factored as (a+b)(a-b)...

Answer 3

Okay, I did that and got 1+cos x, giving me a derivative of -sinx... but the answer is 2...

Answer 4

I have no idea how either...

Answer 5

Help? Anyone? I'm really stuck here... please?

Answer 6

OK, let's do it:\[\lim_{x \rightarrow 0}\frac{ \sin^2x }{ 1-\cos x }=\lim_{x \rightarrow 0}\frac{ 1-\cos^2x }{ 1-\cos x }=\lim_{x \rightarrow 0}\frac{ (1-\cos x)(1+\cos x) }{ 1-\cos x }\]Now the factor 1-cos x cancels out, so we have only left:
\[\lim_{x \rightarrow 0}(1+\cos x)=2\]

Answer 7

See? no need to differentiate, the factor that caused problems (giving 0/0) has been cancelled out, so to calculate the limit, you can just take 1+ cos 0 =1+1 =2.

Answer 8

OH!!!!! Derp... wow I feeel like a noob... THANK YOU SO MUCH ZeHanz! I did the wrong thing! That makes a heck of a lot more sense now! I kept trying to use the quotient rule! >_<