logical solution of trigonometry limit

Question

Most colculators say that $\lim_{x\to 0} \frac {\cos x^\frac {2}{3} - 1}{x} = 0$, but how do we get to this? I tried to solve it the same way as we interpret $\lim_{x\to 0} \frac {\cos x - 1}{x} = \lim_{x\to 0} \frac {\cos^2 x - 1}{x(\cos x+1)}=-\lim_{x\to 0} \frac {\sin^2 x}{x(\cos x+1)}=-\lim_{x\to 0} \frac {\sin x}{x} \sin x \frac {1}{\cos x + 1}=0$, but in the end i got $-\infty * 0 * 1/2$. Looking for ideas guys.

Answer 1

You can use the same way. $$\lim_{x \to 0} \frac{\cos x^{\frac{2}{3}}-1}{x} = \lim_{x \to 0} \frac{\cos x^{\frac{2}{3}}-1}{x}\cdot \frac{\cos x^{\frac{2}{3}}+1}{\cos x^{\frac{2}{3}}+1}$$

$$\implies \lim_{x \to 0} \frac{\cos^2 x^{\frac{2}{3}}-1}{x\big(\cos x^{\frac{2}{3}}{+1}\big)} = -\lim_{x \to 0} \frac{\sin^2 x^{\frac{2}{3}}}{x\big(\cos x^{\frac{2}{3}}{+1}\big)} = -\lim_{x \to 0} \frac{\sin x^{\frac{2}{3}}}{x} \cdot \lim_{x\to 0} \frac{\sin x^{\frac{2}{3}}}{\cos x^{\frac{2}{3}}+1}$$

The latter limit tends to $0$.

Answer 2

hint

$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$

$$\cos^\frac 23(x)-1=\frac{-\sin^2(x)}{\cos^\frac 43(x) +\cos^\frac 23(x)+1}$$

the limit is $\frac 03$=zero.