Calculate the limit or prove it doesn't exist: $\lim_{x\to 0}\frac{\sin(x)}{1/\cos(x) - 1}$ This is the limit:
$$ \lim_{x\to 0}\frac{\sin(x)}{\frac{1}{\cos(x)} - 1}\ $$
Now I'm not sure if I'm allowed to use arithmetic of limits, as I don't know if the limit exist.
I don't understand why I can't just say that:
$$ \lim_{x\to 0} \sin(x) = 0 $$ and therefore the upper limit is also 0 (regardless of the denominator..
Thank you
 A: Write as 
$$\cos x\frac{\sin x}{x}\frac{x^2}{1-\cos x}\frac{1}{x}$$
All factors have a finite limit except the last, which as limit $\infty$, so it does not exist.
A: Hint:
$${\sin x\over{1\over\cos x}-1}={\sin x\cos x\over1-\cos x}={\sin x\cos x (1+\cos x)\over1-\cos^2x}={\cos x(1+\cos x)\over\sin x}$$
A: Multiply through by $\cos x$ $(\text{NB }\neq 0)$ to clear fractions and obtain $$\frac {\sin x\cos x}{1-\cos x}=\frac {2\sin {\frac x2}\cos {\frac x2}\cos x}{2\sin^2 {\frac x2}}=\frac {\cos {\frac x2}\cos x}{\sin {\frac x2}}$$And now the behaviour should be obvious.
A: Notice that
$$\lim_{x\to0}\frac xx=1$$
This should be fairly obvious since $x/x=1$.
But by your logic, this limit should be $0$.
But you forget that the denominator is also $0$, so it is actually an indeterminate form.  If the denominator were anything else, then your reasoning would be right.
You should look towards the other answers to find more suitable methods for evaluating these types of limits.
A: HINT:
$$\frac{\sin(x)}{\frac1{\cos(x)}-1}=\frac{\frac12\sin(2x)}{1-\cos(x)}=\frac{\frac12\sin(2x)}{2\sin^2(x/2)}\sim \frac{x}{x^2/2}$$
