Evaluating limit without L'Hospital's I would like to evaluate the limit $$\lim_{x\to0}\frac{2x\sin(x)}{1-\cos(x)}$$ It can be done with L'Hospital, but I would like to know if it is possible to do so without L'Hospital. I tried multiplying through by the conjugate of the denominator, which leads me to $$\lim_{x\to0}\frac{2x(1+\cos(x))}{\sin(x)}$$ which is not very helpful, because it is still $0/0$. Any help would be much appreciated, thanks!
From the later form you can use $\lim_{x\to 0} \frac{\sin x}{x} = 1$ directly with the product rule to show that the limit is $4$. This identity can be proven without L'Hopital so it's valid to use it. A lot of limits involving trigonometric functions can be solved this way
Hint
$$1-\cos(x)=2\sin^2\left(\frac{x}{2}\right),$$
and $$\lim_{x\to 0}\frac{\sin(x)}{x}=1.$$