Limit $\lim_{x \to 0}\frac{\sin(x^2-x)}{x^2-\tan x}$ without l'Hospital

Question

$$\lim_{x \to 0}\frac{\sin(x^2-x)}{x^2-\tan x}$$

I could not solve this without l'hospital.

Answer 1

$$\lim_{x\rightarrow0}\frac{\sin\left(x^{2}-x\right)}{x^{2}-\tan x}=\lim_{x\rightarrow0}\frac{\sin\left(x^{2}-x\right)}{x^{2}-x}\frac{x^{2}-x}{x^{2}-\frac{\sin x}{\cos x}}=\lim_{x\rightarrow0}\frac{\sin\left(x^{2}-x\right)}{x^{2}-x}\cos x\frac{x-1}{x\cos x-\frac{\sin x}{x}}=1$$

Answer 2

Hint : Use the taylor series of the numerator and the denominator.