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Exercise 9.1 · Q18

Q.Sketch the graph of a function ff that satisfies the given values. (i)
f(0)f(0) is undefined lim⁡x→0f(x)=4\displaystyle\lim_{x\to0}f(x)=4 f(2)=6f(2)=6 lim⁡x→2f(x)=3\displaystyle\lim_{x\to2}f(x)=3 (ii)
f(−2)=0f(-2)=0 f(2)=0f(2)=0 lim⁡x→−2f(x)=0\displaystyle\lim_{x\to-2}f(x)=0 lim⁡x→2f(x)\displaystyle\lim_{x\to2}f(x) does not exist.

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Step 1. Restate the conditions for part (i). f(0)f(0) undefined (a hole at x=0x=0); lim⁡x→0f(x)=4\displaystyle\lim_{x\to0}f(x)=4 (both sides of the hole rise to height 44); f(2)=6f(2)=6 (a plotted, solid point at (2,6)(2,6)); lim⁡x→2f(x)=3\displaystyle\lim_{x\to2}f(x)=3 (both sides near x=2x=2 approach height 33, different from the plotted point). These four conditions are mutually consistent: a hole and a plotted-but-mismatched point are two independent, unrelated features that can coexist on the same curve.

Step 2. Describe a valid sketch for part (i). Draw a smooth curve that rises to touch height 44 as x→0x\to0 from both sides but leave an open circle exactly at (0,4)(0,4) (undefined there). Separately, near x=2x=2 draw the curve approaching height 33 from both sides (open circle at (2,3)(2,3)), but mark a solid dot at (2,6)(2,6), disconnected from the curve — this is a classic removable-type discontinuity where the function's actual value disagrees with its limit.

Step 3. Restate the conditions for part (ii). f(−2)=0f(-2)=0 and lim⁡x→−2f(x)=0\displaystyle\lim_{x\to-2}f(x)=0 (value equals limit — continuous there); f(2)=0f(2)=0 but lim⁡x→2f(x)\displaystyle\lim_{x\to2}f(x) does not exist (the plotted value is defined, but the two one-sided limits disagree). These are consistent: continuity at one point and a jump discontinuity at another are independent local behaviours. …

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