- #1
issacnewton
- 1,026
- 36
Homework Statement
Prove that
$$ \lim_{x\to 0} \sqrt{x^3+x^2}\; \sin\left(\frac{\pi}{x}\right) = 0 $$
using Sandwich theorem
Homework Equations
Sandwich Theorem
The Attempt at a Solution
Now we know that sine function takes values between -1 and 1. ## -1 \leqslant \sin\left(\frac{\pi}{x}\right) \leqslant 1 ##. So we can multiply this by ## \sqrt{x^3+x^2} ## on both sides. But I want to show that ## \sqrt{x^3+x^2} \geqslant 0## near ## x = 0##. So I am stuck at this point. Any guidance will help.
Thanks