# Using the Squeeze Theorem to evaluate $\lim_{x\to0}\sqrt[3]{x}\cos(\ln(x^4))$

So I have this limit here:

$$\lim_{x\to0}\sqrt[3]{x}\cos(\ln(x^4))$$

Would this just be a simple application of the Squeeze Theorem? (I can't use L'hopital or Taylor polynomials.)

$$-\sqrt[3]{x}\leq\sqrt[3]{x}\cos(\ln(x^4))\leq\sqrt[3]{x}$$

The limit from both sides is $0$, so the middle limit must also be $0$. Is that right?

• Yes. You know this because cos(x) is always bounded by -1 and 1. – user2825632 Sep 10 '16 at 22:53

Yes, you are right. Since $$-1\leq \cos(\ln(x^4)) \leq 1$$ for all $x$ ($\neq 0$) you indeed have $$-\sqrt[3]{x}\leq\sqrt[3]{x}\cos(\ln(x^4))\leq\sqrt[3]{x}$$
and so $$\lim_{x\to 0} -\sqrt[3]{x} = \lim_{x\to 0} \sqrt[3]{x} = 0$$ by the Squeeze Theorem the function in the middle will also have limit $0$: $$\lim_{x\to 0}\sqrt[3]{x}\cos(\ln(x^4)) = 0.$$