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Existence of a limit for an oscillating function

Consider the function f(x)=sin⁡(1x)f(x) = \sin\left(\frac{1}{x}\right), which is defined for all x≠0x \neq 0. As xx gets closer and closer to 00, the value of 1x\frac{1}{x} grows without bound, causing the sine function to oscillate faster and faster between −1-1 and 11.

The graph of f(x)=sin⁡(1x)f(x) = \sin\left(\frac{1}{x}\right) near x=0x = 0 is shown below.

Existence of a limit for an oscillating function graph

Using the graph (and/or by picking a sequence of xx-values approaching 00 from the right, such as x=1π/2,13π/2,15π/2,…x = \frac{1}{\pi/2}, \frac{1}{3\pi/2}, \frac{1}{5\pi/2}, \ldots), determine whether lim⁡x→0+sin⁡(1x)\displaystyle\lim_{x \to 0^+} \sin\left(\frac{1}{x}\right) exists. If it exists, give its value; if it does not exist, state why not.

Hint

Think about what happens to 1/x1/x as x→0+x \to 0^+, and how sin⁡\sin behaves as its input grows without bound. Does f(x)f(x) settle down and approach a single number, or does it keep repeating a pattern of values forever?

Step-by-step solution

As x→0+x \to 0^+, the quantity 1x\frac{1}{x} increases without bound, so we are evaluating sin⁡\sin at inputs that sweep through larger and larger angles. Since sin⁡(θ)\sin(\theta) cycles repeatedly through all values between −1-1 and 11 as θ→∞\theta \to \infty, the outputs of f(x)=sin⁡(1/x)f(x) = \sin(1/x) do not settle toward any single number as x→0+x \to 0^+.

For instance, choosing x=1(π/2)+2nπx = \frac{1}{(\pi/2) + 2n\pi} gives f(x)=1f(x) = 1 for every integer nn, while choosing x=1(3π/2)+2nπx = \frac{1}{(3\pi/2)+2n\pi} gives f(x)=−1f(x) = -1. Since we can find xx-values arbitrarily close to 0+0^+ where f(x)=1f(x) = 1 and other xx-values arbitrarily close to 0+0^+ where f(x)=−1f(x) = -1, the function does not approach one fixed value. Therefore lim⁡x→0+sin⁡(1/x)\lim_{x \to 0^+} \sin(1/x) does not exist due to oscillation.

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