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Exercise 9.5 · Q15

Q.State how continuity is destroyed at x=x0x=x_0 for each of the following graphs (Fig. 9.38-9.41). (a) A curve drawn for x<x0x<x_0 ends at a solid (filled) point directly above x0x_0; a second branch starts at an open circle a little lower (directly below that solid point) and continues for x>x0x>x_0. (b) A curve drawn for x<x0x<x_0 approaches an open circle above x0x_0 from the left; a separate branch (of the same underlying curve) resumes just to the right of x0x_0, a little higher up, with no point plotted at x0x_0 itself. (c) A curve has a vertical asymptote at x=x0x=x_0 (dashed line): it plunges to −∞-\infty as x→x0−x\to x_0^- and rises up from +∞+\infty as x→x0+x\to x_0^+. (d) A curve drawn for x<x0x<x_0 approaches an open circle above x0x_0; a second branch starts at a solid point a little lower (directly below that open circle) and continues upward for x>x0x>x_0.

Four small graphs (a)-(d), each showing continuity destroyed at x = x0: (a) JUMP — a curve for x<x0 ends at a SOLID point above x0, a — Mathematics question
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For each figure, identify which of the three continuity conditions — f(x0)f(x_0) defined, the limit exists, the limit equals f(x0)f(x_0) — fails, and classify the discontinuity using the standard vocabulary: removable (limit exists, but f(x0)f(x_0) is either undefined or doesn't match it), jump (both one-sided limits exist but disagree with each other), or infinite (a one-sided limit is unbounded, so no finite limit exists at all).

Step 1. Fig 9.38 — graph (a). The curve approaches, from the left, a solid filled point directly above x0x_0 — so the left-hand limit exists and equals the height of that solid dot, call it LL. The function's actual value f(x0)f(x_0), however, is plotted lower, at an open circle at x0x_0 itself, from which the right branch continues. So the (two-sided) limit exists and equals LL, but f(x0)≠Lf(x_0)\ne L (it sits at the lower, differently-plotted height). Since the limit exists but disagrees with the function's actual value at the point, this is a removable discontinuity — redefining f(x0)f(x_0) to equal LL would restore continuity.

Step 2. Fig 9.39 — graph (b). The curve is a single smooth unbroken shape on both sides of x0x_0, with only ONE point missing: an open circle sits where the curve would naturally pass directly above x0x_0, but there is no solid dot plotted anywhere at x0x_0. So both the left-hand and right-hand limits exist and agree (it's visibly the same continuous curve on both sides), meaning lim⁡x→x0f(x)\lim_{x\to x_0}f(x) exists — but f(x0)f(x_0) itself is simply undefined (no value assigned at x0x_0). Since the limit exists but f(x0)f(x_0) fails to be defined at all, this is also a removable discontinuity — defining f(x0)f(x_0) to equal that limit would restore continuity.

Step 3. Fig 9.40 — graph (c). The curve has a vertical asymptote at x=x0x=x_0 (shown as a dashed vertical line): it plunges toward −∞-\infty as x→x0−x\to x_0^- and rises up from +∞+\infty as x→x0+x\to x_0^+. Neither one-sided limit is a finite number — both are unbounded — so lim⁡x→x0f(x)\lim_{x\to x_0}f(x) does not exist in any finite sense. This is an infinite discontinuity: it can never be "fixed" by redefining a single point, since the function is unbounded near x0x_0. …

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