Exercise · ZW-SQUEEZE-THEOREM-001 · ★★☆ Intermediate
What is \(\lim_{x\to 0} x^{2}\sin\!\left(\dfrac{1}{x}\right)\)?
Choose one option.
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- A. 0
Exercise · ZW-SQUEEZE-THEOREM-001 · ★★☆ Intermediate
Choose one option.
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.
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 \(a\).