Answer
Squeeze Theorem
$lim _{x\to 0}\left(\sqrt{x^3+x^2}\right)sin\frac{\pi }{x}=0$
Work Step by Step
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.
