Im having trouble calculating this limit:
$$\lim_{x \to 0} \left(\cot x-\frac{1}{\sin x}\right)$$
I've tried factoring out $\frac{1}{\sin x}$, $\cos x$, $\cot x$ and it doesn't lead me anywhere. I also tried looking at alternative form $\frac{\cos x-1}{\sin x}$, with no luck. I can't use l'Hospitals rule. Can anyone help? I think this is super easy but somehow I'm stuck.
Best Answer
$\frac{(\cos x-1)(\cos x +1)}{\sin x (\cos x+1)}= \frac{\cos^2-1}{\sin x(\cos x +1)}=$
$\frac{-\sin^2 x}{\sin x(\cos x+1)}=\frac{-\sin x}{\cos x +1}$.
And now?