Answer
Squeeze Theorem
$lim _{x\to 0}(x^2cos20\pi x)=0$
Work Step by Step
Squeeze Theorem
If $h\left(x\right)\le f\left(x\right)\le g\left(x\right)$
If $lim _{x\to a}h\left(x\right)=L$ and $lim _{x\to a}g\left(x\right)=L$ then:
$lim _{x\to a}f\left(x\right)=L$
We know $-1\le cos20\pi x\le 1$
Multiply by $x^{2}$:
$-x^2\le x^2 cos20\pi x\le x^2$
Since $lim _{x\to 0}\left(-x^2\right)=0$
$lim _{x\to 0}\left(x^2\right)=0$
By Squeeze Theorem
$lim _{x\to 0}(x^2cos20\pi x)=0$
See graph.
