## Calculus: Early Transcendentals 8th Edition

Squeeze Theorem $lim _{x\to 0}(x^2cos20\pi x)=0$
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.