Limit of $\displaystyle \lim_{x\to 0+}\frac 1x-\frac{\cos x}{\sin x}=0$ How to best show that 
$$\lim_{x\to 0+}\frac 1x-\frac{\cos x}{\sin x}=0.$$

I've tried to make a fraction out of it, but then I get 0/0, and so I take the derivative using L'Hopitals, and it just gets messier and messier. 

Would just like a hint.
$$\begin{align}\lim_{x\to 0+}\frac 1x-\frac{\cos x}{\sin x}&amp;=\lim_{x\to 0+}\frac{\sin x-x\cos x}{x\sin x}\\&amp;=\lim_{x\to 0+}\frac{(\sin x-x\cos x)'}{(x\sin x)'}\\&amp;=\lim_{x\to 0+}\frac{\cos x-(\cos x-x\sin x)}{\sin x+x \cos x}\\&amp;=\lim_{x\to 0+}\frac{\sin x}{(\sin x/x)+\cos x}\\&amp;=\frac{0}{1+1}\\&amp;=0.\end{align}$$
Here, note that $\lim_{x\to 0}\frac{\sin x}{x}=1.$
$$\lim_{x \to 0}\frac{1}{x}-\frac{\cos (x)}{\sin(x)}=\lim_{x \to 0}\frac{1}{\sin (x)}-\frac{\cos (x)}{\sin(x)}=\frac{1-\cos (x)}{\sin(x)}=\lim_{x \to 0} \frac{\ \sin(x)}{\cos(x)}=\frac{0}{1}=0$$

Here we used L'Hopital (after the third $=$ sign) and we used the identity $\lim_{x \to 0} \frac{ x}{\sin (x)}=1$