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Exercise 8.6 · Q1

Q.Study the graph given below: the graph shows a decreasing straight line for x<3x < 3 that approaches an open (unfilled) circle at the point (3,−1)(3, -1) (this value is not attained), and for x≥3x \geq 3 a horizontal ray at y=2y = 2 starting with a filled dot at (3,2)(3, 2).

(i) What y-value is the function approaching as xx approaches 3 from the left?
(ii) What y-value is the function approaching as xx approaches 3 from the right?
(iii) What (if any) is the actual y-value at x=3x=3? What can you conclude about the function?
A graph on x-y axes: a DECREASING straight line for x<3 that approaches an OPEN (unfilled) circle at the point (3, -1) (value not — Applied Mathematics question
Figure
Chandigarh CbseNCERTSubjective· 3mImportance★★★★★est
60% · 12/20 Questions
✓ Free question

Read the one-sided limits straight off the described graph and compare them with the actual function value at x=3x=3 to test continuity.

lim⁡x→a−f(x)=left-hand limit,lim⁡x→a+f(x)=right-hand limit\lim_{x\to a^-} f(x) = \text{left-hand limit}, \qquad \lim_{x\to a^+} f(x) = \text{right-hand limit}

ff is continuous at x=ax=a iff all three of the following are equal: the left-hand limit, the right-hand limit, and f(a)f(a) itself.

  1. (i) Left-hand limit (x→3−x\to 3^-). For x<3x<3 the graph is a decreasing straight line heading toward the open (unfilled) circle at (3,−1)(3,-1) — the curve gets arbitrarily close to y=−1y=-1 as xx approaches 33 from the left, even though that exact point is not part of the graph.

lim⁡x→3−f(x)=−1\lim_{x\to 3^-} f(x) = -1

  1. (ii) Right-hand limit (x→3+x\to 3^+). For x≥3x\ge 3 the graph is the horizontal ray y=2y=2, starting at the filled dot (3,2)(3,2). As xx approaches 33 from the right, the function value stays at 22.

lim⁡x→3+f(x)=2\lim_{x\to 3^+} f(x) = 2

  1. (iii) Actual value at x=3x=3 and conclusion. The filled dot at (3,2)(3,2) tells us the function is defined at x=3x=3, with

f(3)=2f(3) = 2

Compare the two one-sided limits:

lim⁡x→3−f(x)=−1≠ 2=lim⁡x→3+f(x)\lim_{x\to 3^-} f(x) = -1 \ne \ 2 = \lim_{x\to 3^+} f(x)

Since the left- and right-hand limits disagree, the two-sided limit lim⁡x→3f(x)\lim_{x\to 3} f(x) does not exist. Because continuity at a point requires the two-sided limit to exist and equal f(3)f(3), and here it fails at the very first condition, ff has a jump discontinuity at x=3x=3 — despite f(3)=2f(3)=2 being a perfectly defined value.

  1. Self-check. A jump discontinuity is exactly this pattern: the graph "teleports" from one yy-level (here −1-1, unattained) to another (here 22, attained) at the break point — consistent with an open circle on one branch and a filled dot on the other.
✓Final answer

(i) −1-1. (ii) 22. (iii) f(3)=2f(3)=2, but the one-sided limits differ (−1≠2-1\ne 2), so lim⁡x→3f(x)\lim_{x\to3}f(x) does not exist and ff is discontinuous at x=3x=3 (a jump discontinuity).

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