Limit $\lim_{x \to 0} \left(\cot x-\frac{1}{\sin x}\right)$ 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. 
 A: Using standard limits
$$\cot x - \frac 1{\sin x} = x\frac{\cos x - 1}{x^2}\frac{x}{\sin x}\stackrel{x\to 0}{\longrightarrow}0\cdot \frac{-1}{2}\cdot 1=0$$
A: $\frac{(\cos x-1)(\cos x +1)}{\sin x (\cos x+1)}= \frac{\cos^2-1}{\sin x(\cos x +1)}=$
$\frac{-\sin^2 x}{\sin x(\cos x+1)}=\frac{-\sin x}{\cos x +1}$.
And now?
A: $$\lim_{x \to 0} \left(\cot x-\frac{1}{\sin x}\right)=\lim_{x\to 0}\frac{\cos x-1}{\sin x}=\lim_{x\to 0}\frac{-2\sin^2\left(\frac{x}{2}\right)}{2\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}=0.$$
A: $$\cot x-\frac{1}{\sin x}=\frac{\cos x}{\sin x}-\frac{1}{\sin x}=\frac{\cos x-1}{\sin x}$$
The Taylor series expansion for $\sin(x)$ and $\cos(x)$ at $x=0$ is
$$\sin(x)=x-\frac{x^3}{3!}+\mathcal{O}(x^5)$$
$$\cos(x)=1-\frac{x^2}{2!}+\mathcal{O}(x^4)$$
therefore
$$\cos(x)-1=-\frac{x^2}{2!}+\mathcal{O}(x^4)$$
hence
$$\frac{\cos x-1}{\sin x}=\frac{-\frac{x^2}{2!}+\mathcal{O}(x^4)}{x-\frac{x^3}{3!}+\mathcal{O}(x^5)}=\frac{-\frac{x}{2!}+\mathcal{O}(x^3)}{1-\frac{x^2}{3!}+\mathcal{O}(x^4)}$$
then $\frac{\cos x-1}{\sin x} \to \frac{0}{1}=0$ as $x\to 0$.
