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One-sided limits of x*sin(1/x) at x = 0

Consider the function f(x)=xsin⁡(1x)f(x) = x \sin\left(\dfrac{1}{x}\right), defined for all x≠0x \neq 0.

Since −1≤sin⁡(1x)≤1-1 \le \sin\left(\dfrac{1}{x}\right) \le 1 for all x≠0x \neq 0, we have the bounds

−∣x∣≤xsin⁡(1x)≤∣x∣.-|x| \le x\sin\left(\frac{1}{x}\right) \le |x|.

Using this bound (or by examining values of f(x)f(x) for xx close to 00 from each side), determine the value of the left-hand limit lim⁡x→0−xsin⁡(1x)\displaystyle\lim_{x\to 0^-} x\sin\left(\frac{1}{x}\right) and the value of the right-hand limit lim⁡x→0+xsin⁡(1x)\displaystyle\lim_{x\to 0^+} x\sin\left(\frac{1}{x}\right). Note that sin⁡(1/x)\sin(1/x) itself oscillates wildly as x→0x \to 0 and has no limit — your answer should explain why multiplying by xx changes this.

One-sided limits of x*sin(1/x) at x = 0 graph

What are the values of lim⁡x→0−f(x)\displaystyle\lim_{x\to 0^-} f(x) and lim⁡x→0+f(x)\displaystyle\lim_{x\to 0^+} f(x)?

Hint

Since ∣xsin⁡(1/x)∣≤∣x∣|x\sin(1/x)| \le |x|, both bounding functions ∣x∣|x| and −∣x∣-|x| approach the same value as x→0x \to 0 from either side. Use this squeeze to pin down each one-sided limit, even though sin⁡(1/x)\sin(1/x) alone oscillates without settling.

Step-by-step solution

For xx near 00 (from either side), the factor sin⁡(1/x)\sin(1/x) always stays between −1-1 and 11, so −∣x∣≤xsin⁡(1/x)≤∣x∣-|x| \le x\sin(1/x) \le |x|. As x→0−x \to 0^-, both −∣x∣-|x| and ∣x∣|x| approach 00, so by the Squeeze Theorem lim⁡x→0−xsin⁡(1/x)=0\lim_{x\to 0^-} x\sin(1/x) = 0 as well.

The same squeezing argument applies as x→0+x \to 0^+: both bounding functions −∣x∣-|x| and ∣x∣|x| go to 00, forcing f(x)f(x) to be trapped arbitrarily close to 00 too. So lim⁡x→0+xsin⁡(1/x)=0\lim_{x\to 0^+} x\sin(1/x) = 0 as well. Even though sin⁡(1/x)\sin(1/x) oscillates without limit near 00, the shrinking factor xx forces the product to have a well-defined one-sided (and two-sided) limit of 00.

Role of the shrinking factor in the squeeze

Hint
Step-by-step solution

The x part forces both one-sided limits to zero. The sine of one over x factor oscillates endlessly and never settles on its own. But since it stays trapped between negative one and one, multiplying by x, which shrinks toward zero, squeezes the whole product down to zero from both sides.

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