0
$\begingroup$

$$\lim_{x\to 0}x^4\sin\left(\frac{\pi}{3x^2}\right)$$

The answer is $0$, but when I plug in the limit to this question, the denominator will be $0$, which will make this undefined. So, I don't get how does this limit equal to $0$.

Please help, thank you!

$\endgroup$
1
  • 3
    $\begingroup$ The limit concerns the behaviour near but not at $0$. What you've discovered is that $x=0$ is a removable discontinuity. $\endgroup$
    – J.G.
    Commented Sep 26, 2021 at 21:52

2 Answers 2

4
$\begingroup$

Recall that when we consider $x\to 0$ we never reach the value $x=0$ therefore the expression is always well defined, moreover we have $|\sin \theta|\le 1$ and from here we can easily conclude by squeeze theorem using that

$$\left|x^4\sin\left(\frac{\pi}{3x^2}\right)\right| \le x^4 \cdot 1=x^4$$

$\endgroup$
0
$\begingroup$

Note that $\sin\left(\frac{\pi}{3x^2}\right)$ is a bounded function and $\lim_{x \to 0} x^4=0$. By a theorem in Real Analysis, this limit must be equal to $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .