Calculus Limit problem confusion

[wilxing]

Hi there I am doing homework for a calculus class and I got stuck on a limit problem. The original problem is "evaluate limx>0 of (1-2cos(x)+cos^2(x))/(xsin(x)). I simplified this down to: 0 * limx>0 of (cos(x)-1)/sin(x). I thought that I could say that the answer is just 0 because 0 times anything is still 0, but then I thought that I needed to prove that the limit on the right exists and is not undefined but I couldn't figure out how to solve/prove that. How should i go about this?

 

[Ishuda]

Hi there I am doing homework for a calculus class and I got stuck on a limit problem. The original problem is "evaluate limx>0 of (1-2cos(x)+cos^2(x))/(xsin(x)). I simplified this down to: 0 * limx>0 of (cos(x)-1)/sin(x). I thought that I could say that the answer is just 0 because 0 times anything is still 0, but then I thought that I needed to prove that the limit on the right exists and is not undefined but I couldn't figure out how to solve/prove that. How should i go about this?

What tools do you have at your disposal? L'Hopital's rule? The Taylor (actually McClaurain) series expansion of the sine and cosine functions?

From what you said, it looks like you wrote the problem as
\(\displaystyle \lim_{x->0}\frac{(cos(x) - 1)^2}{x sin(x)} = \lim_{x->0}\frac{cos(x)-1}{x} * \frac{cos(x) -1}{sin(x)} = 0 * \lim_{x->0}\frac{cos(x) -1}{sin(x)}\)
If so, how did you get the first part of that went to zero as x goes to zero? Why can't you use the same method to determine the second part?