2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Yes. You know this because cos(x) is always bounded by -1 and 1. $\endgroup$ – user2825632 Sep 10 '16 at 22:53
2
$\begingroup$

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 $0$: $$ \lim_{x\to 0}\sqrt[3]{x}\cos(\ln(x^4)) = 0. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.