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One-sided limits and existence of a two-sided limit at a piecewise breakpoint

Consider the piecewise function r(x)={x2−4if x<0x+1if x≥0r(x) = \begin{cases} x^2 - 4 & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases}

Determine the left-hand limit lim⁡x→0−r(x)\lim_{x \to 0^-} r(x) and the right-hand limit lim⁡x→0+r(x)\lim_{x \to 0^+} r(x). Based on these, does lim⁡x→0r(x)\lim_{x \to 0} r(x) exist? If it exists, give its value; if not, explain why not.

One-sided limits and existence of a two-sided limit at a piecewise breakpoint graph
Hint

Use the formula that applies on each side of 0 separately: for xx slightly less than 0, use x2−4x^2-4; for xx slightly greater than or equal to 0, use x+1x+1. Compare the two resulting values.

Step-by-step solution

For xx approaching 00 from the left, x<0x < 0, so r(x)=x2−4r(x) = x^2 - 4. As x→0−x \to 0^-, x2→0x^2 \to 0, so r(x)→0−4=−4r(x) \to 0 - 4 = -4. Thus lim⁡x→0−r(x)=−4\lim_{x \to 0^-} r(x) = -4.

For xx approaching 00 from the right, x≥0x \geq 0, so r(x)=x+1r(x) = x + 1. As x→0+x \to 0^+, r(x)→0+1=1r(x) \to 0 + 1 = 1. Thus lim⁡x→0+r(x)=1\lim_{x \to 0^+} r(x) = 1. Since the left-hand limit (−4-4) does not equal the right-hand limit (11), the two-sided limit lim⁡x→0r(x)\lim_{x \to 0} r(x) does not exist.

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