$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)$ without L'Hospitals rule? $$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)=?$$
How to solve it without using L'Hospitals rule?

Use equivalents and Taylor's formula at order $2$: we know that


*

*$\cos x=1-\dfrac{x^2}2+o(x^2)$,

*$\sqrt{1+u}=1+\dfrac u2+o(u)$,

*$\sqrt[3]{1+u}=1+\dfrac u3+o(u)$
Thus the numerator gives, by composition:
$$\sqrt[2]{\cos x}-\sqrt[3]{\cos x}=\Bigl(1-\frac{x^2}4\Bigr)-\Bigl(1-\frac{x^2}6\Bigr)+o(x^2)=-\frac{x^2}{12}+o(x^2)$$
so that $\;\sqrt[2]{\cos x}-\sqrt[3]{\cos x}\sim_0 -\dfrac{x^2}{12}$, and
$$\frac{\sqrt[2]{\cos x}-\sqrt[3]{\cos x}}{\sin^2 x}\sim_0 -\dfrac{x^2}{12 x^2}=-\frac1{12}.$$

