Answer
See explanations.
Work Step by Step
Step 1. Using the figure provided by the Exercise, the function is sandwiched between two parabolas $y=\pm x^2$ or $−x^2≤x^2sin\frac{1}{x}≤x^2$. This is because the function $sin\frac{1}{x}$ oscillates within $[−1,1]$
Step 2. Find the limits of the two end functions: $\lim_{x\to0}(−x^2)=0$ and $\lim_{x\to0}(x^2)=0$
Step 3. Based on the Sandwich Theorem, we conclude that $\lim_{x\to0}(x^2sin\frac{1}{x})=0$
