Can you please help me solve:
$$\lim_{x \rightarrow 0} \frac{1- \cos x}{x \sin x}$$
Every time I try to calculate it I find another solution and before I get used to bad habits, I'd like to see how it can be solved right, so I'll know how to approach trigonometric limits.
I tried to convert $\cos x$ to $\sin x$ by $\pi -x$, but I think it's wrong. Should I use another identity?
Best Answer
multiplying numerator and denominator by$$1+\cos(x)$$ we obtain $$\frac{1-\cos(x)^2}{x\sin(x)(1+\cos(x))}=\frac{\sin(x)^2}{x\sin(x)(1+\cos(x))}=\frac{\sin(x)}{x}\cdot \frac{1}{1+\cos(x)}$$