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One-sided limits and existence of the two-sided limit at x = pi/2

Consider the piecewise function f(x)={sin⁡(x)x<π21x≥π2f(x) = \begin{cases} \sin(x) & x < \dfrac{\pi}{2} \\ 1 & x \ge \dfrac{\pi}{2} \end{cases}

Using the definition of f(x)f(x) given above, find lim⁡x→π/2−f(x)\displaystyle\lim_{x \to \pi/2^-} f(x) and lim⁡x→π/2+f(x)\displaystyle\lim_{x \to \pi/2^+} f(x). Then state whether lim⁡x→π/2f(x)\displaystyle\lim_{x \to \pi/2} f(x) exists, and if it does, give its value.

One-sided limits and existence of the two-sided limit at x = pi/2 graph
Hint

Evaluate the left-hand limit using the formula that applies for xx slightly less than π/2\pi/2, and the right-hand limit using the formula that applies for xx slightly greater than or equal to π/2\pi/2. Compare the two results to decide whether the two-sided limit exists.

Step-by-step solution

For xx approaching π/2\pi/2 from the left, x<π/2x < \pi/2, so f(x)=sin⁡(x)f(x) = \sin(x) near there. As x→π/2−x \to \pi/2^-, sin⁡(x)→sin⁡(π/2)=1\sin(x) \to \sin(\pi/2) = 1, so lim⁡x→π/2−f(x)=1\lim_{x \to \pi/2^-} f(x) = 1.

For xx approaching π/2\pi/2 from the right, x≥π/2x \ge \pi/2, so f(x)=1f(x) = 1 identically near there, giving lim⁡x→π/2+f(x)=1\lim_{x \to \pi/2^+} f(x) = 1. Since both one-sided limits equal 11, the two-sided limit exists and lim⁡x→π/2f(x)=1\lim_{x \to \pi/2} f(x) = 1.

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