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Squeeze theorem applied to sqrt(x) sin(1/x) as x -> 0+

Consider the function f(x)=xsin⁡(1x)f(x) = \sqrt{x}\sin\left(\dfrac{1}{x}\right) for x>0x > 0.

Use the Squeeze Theorem to evaluate lim⁡x→0+xsin⁡(1x).\lim_{x \to 0^+} \sqrt{x}\sin\left(\frac{1}{x}\right). Start by writing a chain of inequalities that bounds xsin⁡(1/x)\sqrt{x}\sin(1/x) above and below using the fact that sin⁡\sin is always between −1-1 and 11, then take the limit of the bounding functions to determine the limit of f(x)f(x). State the value of the limit.

Hint

Since −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 for all x≠0x \neq 0, multiply this inequality through by x\sqrt{x} (which is positive for x>0x>0, so the inequality direction is preserved), then find the limits of the resulting lower and upper bounding functions as x→0+x \to 0^+.

Step-by-step solution

Since sin⁡(1/x)\sin(1/x) always satisfies −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 for x≠0x \neq 0, multiply every part of this inequality by x\sqrt{x}, which is positive for x>0x > 0. This preserves the inequality direction and gives −x≤xsin⁡(1x)≤x.-\sqrt{x} \le \sqrt{x}\sin\left(\frac{1}{x}\right) \le \sqrt{x}.

Now take the limit of the outer bounding functions as x→0+x \to 0^+: both lim⁡x→0+(−x)=0\lim_{x\to 0^+}(-\sqrt{x}) = 0 and lim⁡x→0+x=0\lim_{x\to 0^+}\sqrt{x} = 0. Since f(x)=xsin⁡(1/x)f(x) = \sqrt{x}\sin(1/x) is squeezed between two functions that both approach 00, the Squeeze Theorem gives lim⁡x→0+xsin⁡(1x)=0.\lim_{x \to 0^+} \sqrt{x}\sin\left(\frac{1}{x}\right) = 0.

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