Hints to solve $\lim_{x \to 0^+}{\frac{\sqrt x +\tan^3x+\sqrt x\sin^2x}{x+x^2\cos x-\tan^2x}}$ without L'Hopital

Question

$$\lim_{x \to 0^+}{\frac{\sqrt x +\tan^3x+\sqrt x\sin^2x}{x+x^2\cos x-\tan^2x}}.$$

I tried divide both terms by $x^2$ but I didn't get anywhere doing this. Can someone give me some hints on how to solve this limit?

Answer 1

HINT

We have that

$$\frac{\sqrt x +\tan^3x+\sqrt x\sin^2x}{x+x^2\cos x-\tan^2x}=\frac{\sqrt x}{x}\frac{1 +\frac{\tan^3x}{\sqrt x}+\sin^2x}{1+x\cos x-\frac{\tan^2x}x}$$