How to solve a limit with a given condition $ f(0)=\frac{1}{3}$? I need to solve the limit,
$$\lim_{x\rightarrow 0} \frac{1-\cos(x)}{f(x)\sin^2(x)}$$
Limit with a given condition such that $ f(0)=\frac{1}{3}$.
I think that to solve the limit is quite simple, but how to substitute for $x$ with the given condition, $ f(0) = \frac{1}{3}$?
I just need some hints to make it clear.
 A: Note that
$$
\begin{align*}
\lim_{x\to 0}\frac{1-\cos x}{f(x)\sin^2(x)}&=\left[\lim_{x\to 0}\frac1{f(x)}\right]\cdot\left[\lim_{x\to 0}\frac{1-\cos x}{\sin^2x}\right]\\
&=\frac1{f(0)}\cdot\lim_{x\to 0}\frac{1-\cos x}{1-\cos^2(x)}\\&=3\lim_{x\to 0}\frac1{1+\cos x}\\&=\frac32.
\end{align*}
$$
A: As Samuel M. A. Luque mentions in the comments, technically we cannot find the limit unless we are also told that $f$ is continuous at $0$, i.e.
$$
\lim_{x \to 0}f(x)=f(0)=\frac{1}{3} \, .
$$
If $f$ is allowed to be discontinuous at $0$, then the limit is indeterminate because it might be that
$$
f(x)=\begin{cases}
1  &\text{ if $x\neq0$} \\
1/3 &\text{ if $x=0$} \, ,
\end{cases}
$$
in which the case the limit would be equal to $1/2$. If instead
$$
f(x)=\lim_{n \to \infty}\left(1-x^2\right)^n-\frac{2}{3}
$$
then $\lim_{x \to 0}f(x)=-\frac{2}{3}$, and so the overall limit would be equal to $-1/3$. So unless we are told more information (i.e. that $f$ is continuous at $0$), it might well be the case that the answer is not $3/2$.
