1st Year Math Chapter 12 Exercise 12.2 Notes: Continuity, Left-Hand and Right-Hand Limits (Punjab Board 2026-27)
Complete concept notes, formulas, and solved examples for ICS/FSc Part 1 Mathematics, Punjab Board (PECTAA), Single National Curriculum 2026-27 session.
Exercise 12.2 builds on the limit concepts from Exercise 12.1, introducing left-hand and right-hand limits, and the formal definition of continuity at a point.
What Does This Exercise Cover?
This exercise teaches how to check whether a limit exists by comparing its left-hand and right-hand approach, and the three specific conditions a function must satisfy to be considered continuous at a given point.
Key Concepts
1. Left-Hand and Right-Hand Limits
The left-hand limit, written lim(x→c⁻) f(x), describes the value f(x) approaches as x gets close to c from values less than c. The right-hand limit, lim(x→c⁺) f(x), describes the value f(x) approaches from values greater than c.
2. Existence of a Limit
lim(x→c) f(x) exists and equals L if and only if the left-hand limit and the right-hand limit both exist and are equal to the same value L: LHL = RHL = L.
3. Continuity of a Function at a Point
A function f is continuous at a number c if and only if all three of the following conditions hold:
- f(c) is defined
- lim(x→c) f(x) exists
- lim(x→c) f(x) = f(c)
4. Discontinuity
If any one of the three continuity conditions fails to hold at c, the function f is said to be discontinuous at c.
Solved Examples
Example 1: For f(x) = x + 1 if x < 2, and f(x) = 4 if x ≥ 2, find the LHL and RHL at x = 2, and determine whether the limit exists.
LHL: lim(x→2⁻) (x + 1) = 2 + 1 = 3.
RHL: lim(x→2⁺) 4 = 4.
Since LHL ≠ RHL, the two one-sided limits disagree.
Answer: The limit does not exist at x = 2
Example 2: Discuss the continuity of f(x) = x² if x < 2, f(2) = 5, and f(x) = 3x − 2 if x > 2, at x = 2.
f(2) = 5, so condition (i) is satisfied.
LHL: lim(x→2⁻) x² = 4. RHL: lim(x→2⁺) (3x−2) = 4. Since LHL = RHL = 4, the limit exists (condition ii satisfied).
Compare the limit to f(2): 4 ≠ 5, so condition (iii) fails.
Answer: f is discontinuous at x = 2, since the limit (4) does not equal f(2) (5)
Example 3: Determine whether f(x) = (x² − 4)/(x − 2) is continuous at x = 2.
At x = 2, the denominator becomes 0, so f(2) is undefined.
Since condition (i) already fails, there’s no need to check the other two conditions.
Answer: f is discontinuous at x = 2, since f(2) is undefined
Sample MCQs
1. A function f is continuous at x = c if:
a) f(c) is defined only b) The limit exists only c) f(c) is defined, the limit exists, and they are equal d) f is a polynomial
Answer: c) f(c) is defined, the limit exists, and they are equal
2. The left-hand limit is written as:
a) lim(x→c⁺) f(x) b) lim(x→c⁻) f(x) c) lim(x→c) f(x) d) f(c⁻)
Answer: b) lim(x→c⁻) f(x)
3. A limit exists at x = c if and only if:
a) f(c) is defined b) LHL = RHL c) f is continuous there d) f(c) = 0
Answer: b) LHL = RHL
4. If f(c) is undefined, then f is:
a) Continuous at c b) Discontinuous at c c) Always zero at c d) Undefined everywhere
Answer: b) Discontinuous at c
5. If LHL ≠ RHL at x = c, then:
a) f is continuous at c b) The limit does not exist at c c) f(c) = 0 d) f is a polynomial
Answer: b) The limit does not exist at c
Important Short Questions
- Define the left-hand limit and right-hand limit of a function at a point.
- State the condition for the existence of lim(x→c) f(x) in terms of LHL and RHL.
- State the three conditions required for a function to be continuous at a point.
- Give an example of a function that is discontinuous because it is undefined at a point.
- If LHL = 3 and RHL = 5 at x = c, does the limit exist at x = c? Explain.
Important Long Questions
- For f(x) = x + 1 if x < 2, and f(x) = 4 if x ≥ 2, find the LHL and RHL at x = 2, and determine whether the limit exists.
- Discuss the continuity of f(x) = x² if x < 2, f(2) = 5, and f(x) = 3x − 2 if x > 2, at x = 2.
- Determine whether f(x) = (x² − 4)/(x − 2) is continuous at x = 2, explaining which condition fails.
- Explain the difference between a function having a limit at a point and being continuous at that point.
How to Approach This Exercise Effectively
- For piecewise functions, always compute the LHL and RHL separately before deciding whether the overall limit exists.
- Check the three continuity conditions in order — if f(c) is already undefined, you can stop there without checking the limit at all.
- Remember that a limit existing is a weaker condition than continuity — the limit can exist while the function still fails to be continuous if it doesn’t match f(c).
- Sketch a rough graph of piecewise functions when possible — visualizing a jump or hole at the point in question makes the discontinuity obvious.
- Practice identifying which of the three conditions fails in each discontinuous example, rather than just concluding ‘discontinuous’ without justification.
FAQs
Q: What is the difference between a limit existing and a function being continuous?
A: A limit can exist at a point even if the function is undefined there or its value doesn’t match the limit; continuity requires all three conditions — defined, limit exists, and they’re equal — to hold together.
Q: Can a function have a limit at a point where it isn’t defined?
A: Yes — the limit only describes the function’s behavior near the point, not at the point itself, so it can exist even without f(c) being defined.
Q: What are the three ways a function can fail to be continuous at a point?
A: It can fail because f(c) is undefined, because the limit doesn’t exist (LHL ≠ RHL), or because the limit exists but doesn’t equal f(c).
Q: Why do we need both the left-hand and right-hand limit to determine if a limit exists?
A: Because a function could behave completely differently approaching from the left versus the right, especially in piecewise-defined functions — checking both sides is the only way to confirm a single, consistent limiting value.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 12: Limit and Continuity, Exercise 12.2.