## Calculus: Early Transcendentals 8th Edition

Squeeze Theorem $lim _{x\to 0}\left(\sqrt{x^3+x^2}\right)sin\frac{\pi }{x}=0$
Squeeze Theorem If $h(x)≤f(x)≤g(x)$ 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 $-1\le \frac{sin\pi }{x}\le 1$ Multiply by $\sqrt{x^3+x^2}$ $-\sqrt{x^3+x^2}\le \frac{\sqrt{x^3+x^2}sin\pi }{x}\le \sqrt{x^3+x^2}$ Since $\lim _{x\to 0}-\sqrt{x^3+x^2}=0$ $\lim _{x\to 0}\sqrt{x^3+x^2}=0$ Squeeze Theorem $lim _{x\to 0}\left(\sqrt{x^3+x^2}\right)sin\frac{\pi }{x}=0$ See graph.