Answer
0
Note: The green graph is $f(x)=-\sqrt {x^{3}+x^{2}}$, the red one is $h(x)=\sqrt {x^{3}+x^{2}}$ and the final is $g(x)=\sqrt {x^{3}+x^{2}}sin(\frac{\pi}{x})(x\ne0)$
Work Step by Step
We call $g(x)=\sqrt {x^{3}+x^{2}}sin(\frac{\pi}{x})(x\ne0)$. Squeezing the $g(x)$ by $f(x)=-\sqrt {x^{3}+x^{2}}$ and $h(x)=\sqrt {x^{3}+x^{2}}$, we have:
$f(x) \leq g(x) \leq h(x)$.
We all know that:
$\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to 0} (-\sqrt {x^{3}+x^{2}})=0$
$\lim\limits_{x \to 0} h(x)=\lim\limits_{x \to 0} (\sqrt {x^{3}+x^{2}})=0$.
By that Squeeze Theorem, we can conclude:
$\lim\limits_{x \to 0} g(x)=0$
