Exercise · ZW-SQUEEZE-THEOREM-004 · ★☆☆ Basic

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

Choose one option.

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  • B. \(0\)

Explanation

Since \(|\cos(1/x)|\le 1\), we have \(|x\cos(1/x)|\le |x|\), and \(|x|\to 0\). 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

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