Step 1: Bounding the absolute value directly:
For \(x\neq0\), \(|f(x)-f(0)|=|x^2\sin(1/x)-0|=x^2|\sin(1/x)|\le x^2\), using \(|\sin(1/x)|\le1\).
Step 2: Epsilon-delta style bound:
Given any \(\epsilon>0\), choosing \(\delta=\sqrt{\epsilon}\) ensures that whenever \(|x-0|<\delta\), \(|f(x)-f(0)|\le x^2<\delta^2=\epsilon\).
Step 3: Concluding:
This is precisely the \(\epsilon\text{-}\delta\) definition of \(\displaystyle\lim_{x\to0}f(x)=f(0)\).
Final Answer:
Hence \(\boxed{f\text{ is continuous at }x=0}\) (and elsewhere, as a product of continuous pieces), matching the Squeeze Theorem approach.