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

Question

$$\lim _{x\to 0}\left(\frac{\sqrt[2]{\cos \left(x\right)}-\sqrt[3]{\cos \left(x\right)}}{\sin ^2\left(x\right)}\right)=?$$

How to solve it without using L'Hospitals rule?

Answer 1

With $u=\sqrt[6]{\cos x}\to 1$ the term equals $$\frac{u^3-u^2}{1-u^{12}}=u^2\cdot \frac{u-1}{1-u^{12}}=-u^2\frac1{1+u+u^2+\ldots +u^{11}}\to -\frac1{12}$$

Answer 2

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 $$\frac{\sqrt[2]{\cos x}-\sqrt[3]{\cos x}}{\sin^2 x}\sim_0 -\dfrac{x^2}{12 x^2}=-\frac1{12}.$$

Answer 3

Also $$ \lim_{x\to0}\frac{-\left(\frac{\sqrt{1+(\cos x-1)}-1}{\cos x-1}-\frac{\sqrt[3]{1+(\cos x-1)}-1}{\cos x-1}\right)\frac{1-\cos x}{x^2}}{\left(\frac{\sin x}{x}\right)^2}=\frac{-(\frac{1}{2}-\frac{1}{3})\frac{1}{2}}{1^2}=-\frac{1}{12} $$