Exercise · ZW-SQUEEZE-THEOREM-001 · ★★☆ Intermediate

What is \(\lim_{x\to 0} x^{2}\sin\!\left(\dfrac{1}{x}\right)\)?

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  • A. 0

Explanation

Since \(-1\le \sin(1/x)\le 1\), we have \(-x^{2}\le x^{2}\sin(1/x)\le x^{2}\). Both bounds tend to 0 as x → 0, so by the squeeze theorem the limit is 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

ZW-SQUEEZE-THEOREM-002 →