Exercise · ZW-SQUEEZE-THEOREM-006 · ★★☆ Intermediate

What is \(\lim_{x\to\infty}\dfrac{\sin x}{x}\)?

Choose one option.

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  • D. \(0\)

Explanation

Since \(-1\le\sin x\le 1\), for \(x>0\) we have \(\dfrac{-1}{x}\le\dfrac{\sin x}{x}\le\dfrac{1}{x}\). Both bounds tend to 0, so 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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