Question

Prove that the function defined by \(f(x)=\begin{cases}x^2\sin\dfrac1x, & x\neq0\\0, & x=0\end{cases}\) is continuous.

Hint:

Away from 0, \(f\) is a product of continuous functions; at 0, squeeze \(x^2\sin(1/x)\) between \(-x^2\) and \(x^2\).

Answer 1

Step 1: Understanding the Concept:
For \(x\neq0\), \(f\) is a product/composition of standard continuous functions (\(x^2\) and \(\sin(1/x)\)), so continuity there is automatic. The only point that needs checking is \(x=0\), where the formula changes.

Step 2: Continuity for x ≠ 0:
For \(x\neq0\), \(\sin(1/x)\) is continuous (composition of the continuous \(1/x\) with \(\sin\)), and \(x^2\) is continuous, so their product \(f(x)=x^2\sin(1/x)\) is continuous there.

Step 3: Setting up the limit at x = 0:
We need \(\displaystyle\lim_{x\to0}f(x)=f(0)=0\).

Step 4: Squeezing the limit:
Since \(-1\le\sin(1/x)\le1\) for all \(x\neq0\), we get \(-x^2\le x^2\sin(1/x)\le x^2\). As \(x\to0\), both \(-x^2\to0\) and \(x^2\to0\), so by the Squeeze Theorem, \(\displaystyle\lim_{x\to0}x^2\sin(1/x)=0\).

Final Answer:
Since \(\displaystyle\lim_{x\to0}f(x)=0=f(0)\), \(\boxed{f\text{ is continuous at }x=0}\), and hence continuous everywhere.