Find the limit of $(1-\cos x)/(x\sin x)$ as $x \to 0$ 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?
 A: hint
Combine the two Well-known limits
$$\lim_{x \rightarrow 0} \frac {\sin (x)}{x}=\lim_{x \rightarrow 0}\frac {x}{\sin (x)}=1$$
and
$$\lim_{x \rightarrow 0} \frac {1-\cos (x)}{x^2}=\frac {1}{2} $$
A: $$\frac{1-\cos x}{x\sin x}=\frac{2\sin^2\frac x2}{2x\sin\frac x2\cos\frac x2}=\frac12\frac{\sin\frac x2}{\frac x2}\frac1{\cos\frac x2}.$$
A: Use the fact that$$\frac{1-\cos x}{x\sin x}=\frac{(1-\cos x)(1+\cos x)}{x\sin x(1+\cos x)}=\frac{\sin x}{x(1+\cos x)}.$$
A: We can also use L'Hospital's Rule
$$\lim_{x\to0}\frac{1-\cos x}{x\sin x}=\lim_{x\to0}\frac{\sin x}{x\cos x+\sin x}=\lim_{x\to0}\frac{\cos x}{-x\sin x+2\cos x}=\frac{1}{2}$$
A: multiplying numerator and denominator by$$1+\cos(x)$$ we obtain
$$\frac{1-\cos(x)^2}{x\sin(x)(1+\cos(x))}=\frac{\sin(x)^2}{x\sin(x)(1+\cos(x))}=\frac{\sin(x)}{x}\cdot \frac{1}{1+\cos(x)}$$
A: Whilst the other answers are 'clever', note that these sort of limits can be done automatically $$\frac{1-\cos x}{x\sin x} = \frac{x^2/2 + O(x^4)}{x^2 + O(x^4)} = \frac{1/2 + O(x^2)}{1 + O(x^2)} \stackrel{x \to 0}{\longrightarrow} \frac{1}{2}$$
A: Just taking terms of the fraction in question leads to:
\begin{align}
\frac{1-\cos(x)}{x \, \sin(x)} &= \frac{1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right)}{x^2 \, \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \right)} \\ 
&= \frac{\frac{1}{2!} - \frac{x^2}{4!} + \cdots}{1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots } \\
&= \frac{1}{2!} + \frac{x^2}{4!} + \frac{3 \, x^4}{6!} + \mathcal{O}(x^6)
\end{align}
which leads to
$$\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \frac{1}{2}.$$
Alternatively, L'Hospital's rule applies.
\begin{align}
\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} \to \frac{0}{0}
\end{align}
which leads to
\begin{align}
\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x \, \cos(x) + \sin(x)} \to \frac{0}{0}
\end{align}
and finally becomes
\begin{align}
\lim_{x \to 0} \frac{1 - \cos(x)}{x \, \sin(x)} = \lim_{x \to 0} \frac{\sin(x)}{x \, \cos(x) + \sin(x)} = \lim_{x \to 0} \frac{\cos(x)}{2 \, \cos(x) - x \, \sin(x)} = \frac{\cos(0)}{2 \, \cos(0)} = \frac{1}{2}.
\end{align}
A: $$\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac{1}{2}\implies 1-\cos(x)\sim\frac{1}{2}x^2\mbox{ for }x\to 0$$
$$\lim_{x\to 0}\frac{\sin(x)}{x}=1\implies \sin(x)\sim x\mbox{ for }x\to 0$$
so $$\lim_{x\to 0}\frac{1-\cos(x)}{x\sin(x)}=\lim_{x\to 0}\frac{\frac{1}{2}x^2}{x\cdot x}=\frac{1}{2}$$
