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?

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}
