Question

What is the limit as x -> 0 of the function, f(x)= 1-cosx / sinx

Answer 1

Do you know L'Hopital's rule? If not, there's a way to do it without it.

Answer 2

\[\lim_{x\to0}\frac{1-\cos x}{\sin x}=\frac{0}{0}\]
Apply L'Hopital's rule:
\[\lim_{x\to0}\frac{1-\cos x}{\sin x}=\lim_{x\to0}\frac{\sin x}{\cos x}\]

If you don't know L'Hopital's rule, you can use the fact that \(\sin x\approx x\) for \(x\) near 0, so you have
\[\lim_{x\to0}\frac{1-\cos x}{\sin x}=\lim_{x\to0}\frac{1-\cos x}{x}\]
which is a limit you may have already been exposed to.