Exercise · ZW-SQUEEZE-THEOREM-008 · ★★★ Advanced

Let \(f(x)=x^{2}\sin(1/x)\) for \(x\ne 0\) and \(f(0)=0\). What is \(f'(0)=\lim_{h\to 0}\dfrac{f(h)-f(0)}{h}\)?

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  • B. 0

Explanation

The quotient is \(\dfrac{h^{2}\sin(1/h)}{h}=h\sin(1/h)\). Since \(|h\sin(1/h)|\le |h|\to 0\), the squeeze theorem gives 0.

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Review the concept: Squeeze theorem

If \(g(x)\le f(x)\le h(x)\) near \(a\) and \(g\) and \(h\) both have limit \(L\) at \(a\), then \(f\) also has limit \(L\) at \(a\).

\[g\le f\le h,\ \lim_{x\to a}g=\lim_{x\to a}h=L\ \Rightarrow\ \lim_{x\to a}f=L\]

More practice problems: squeeze theorem

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