A function is **continuous** at a point intuitively when its graph can be drawn without lifting the pen from paper. Consider:
**Example of discontinuity:** f(x) = 1 if x ≤ 0, and f(x) = 2 if x > 0. At x = 0, the left limit is 1 and right limit is 2. They don't match, so the function "jumps" — it's discontinuous at x = 0.
**Example of discontinuity at jump:** f(x) = 1/x when x ≠ 0, and f(0) = 2. Although limits exist on both sides (both equal 1), the function value at x = 0 is 2, which doesn't match the limit. Discontinuous at x = 0.
---
**Definition 1:** A real function f is **continuous at a point c** in its domain if:
**lim[x→c] f(x) = f(c)**
This means three conditions must hold simultaneously:
**Alternative phrasing:** f is continuous at c if:
When f is **not continuous** at c, we call c a **point of discontinuity**.
---
**Definition 2:** A function f is **continuous** (on its domain) if it is continuous at **every point** in its domain.
**Special case for closed intervals [a, b]:** f is continuous on [a, b] if:
Note: If a function's domain is a single point, it is continuous there.
---
**Example 1: Linear function continuity**
Check continuity of f(x) = 2x + 3 at x = 1.
**Example 2: Polynomial continuity**
Show f(x) = x² is continuous at x = 0.
**Example 3: Absolute value function**
Discuss continuity of f(x) = |x| at x = 0.
Since |x| is continuous at x = 0 and at all other points (polynomial behavior away from 0), |x| is a **continuous function** everywhere.
**Example 4: Function with jump discontinuity**
Show f(x) = x + 3 when x < 1, f(1) = 0, f(x) = x - 2 when x > 1 is discontinuous at x = 1.
**Example 5: Constant functions**
f(x) = k (constant) is continuous everywhere.
**Example 6: Greatest integer function**
f(x) = [x] (greatest integer ≤ x) is **discontinuous at every integer**.
At c = 2 (an integer):
At c = 2.5 (non-integer):
---
When analyzing f(x) = 1/x near x = 0:
**Right hand limit:**
**Left hand limit:**
This explains why f(x) = 1/x is **discontinuous at x = 0** (and x = 0 is not even in the domain).
---
**All these are continuous functions:**
**Common discontinuous functions:**
---
If **f and g are continuous at x = c**, then:
1. **f + g is continuous at x = c**
2. **f - g is continuous at x = c**
3. **f · g is continuous at x = c**
4. **f/g is continuous at x = c** (provided g(c) ≠ 0)
**Proof sketch for (f + g):**
**Consequences of Theorem 1:**
---
If **g is continuous at c** and **f is continuous at g(c)**, then **f ∘ g is continuous at c**.
Equivalently: **(f ∘ g)(x) = f(g(x))** is continuous when both component functions are continuous.
**Example:** f(x) = sin(x²) is continuous.
**Example:** f(x) = |1 - x + |x|| is continuous.
---
**Common mistakes to avoid:**
**What examiners test:**
---
---
This completes the continuity portion. The concept of differentiability (next section) builds directly on continuity: **differentiability implies continuity, but continuity does NOT guarantee differentiability.**
Q1. Which of the following functions is continuous at x = 0?
Answer: B — For f(x) = |x|, we have f(0) = 0, lim(x→0−) |x| = 0, and lim(x→0+) |x| = 0, so all three values match.
Q2. A function f(x) = x² + 2x is continuous at x = 1 because:
Answer: C — Continuity requires that the function value equals the limit at that point; here f(1) = 1 + 2 = 3 and the limit also equals 3.
Q3. The function f(x) = 5 (constant function) is continuous at:
Answer: C — For any point c, f(c) = 5 and lim(x→c) 5 = 5, satisfying the continuity condition at every real number.
Q4. For the piecewise function f(x) = {x if x ≤ 1, 3 if x > 1}, at which point is there a jump discontinuity?
Answer: B — At x = 1, lim(x→1−) f(x) = 1 but lim(x→1+) f(x) = 3, and these left and right limits are unequal, creating a jump.
Q5. If f(x) = (x³ - 1)/(x - 1) for x ≠ 1 and f(1) = 3, then f is:
Answer: C — Simplifying: (x³ - 1)/(x - 1) = x² + x + 1, so lim(x→1) (x² + x + 1) = 3 = f(1), making it continuous.
Q6. Which statement about continuity is INCORRECT?
Answer: B — If f(c) is undefined, the function cannot be continuous at c because continuity requires f(c) to be defined and equal the limit.
Q7. Consider f(x) = |x - 2|. At x = 2, the function is continuous. Which of the following justifies this?
Answer: D — All three reasons are true: f(2) = 0, both one-sided limits equal 0, and |x - 2| is a standard continuous function type.
Q8. The function f(x) = (x² - 4)/(x - 2) for x ≠ 2 has a removable discontinuity at x = 2 because:
Answer: B — The limit lim(x→2) (x + 2) = 4 exists even though f(2) is undefined, making this a removable discontinuity that could be 'fixed' by redefining f(2) = 4.
Q9. Assertion (A): The function f(x) = x³ + x² - 1 is continuous at x = 0. Reason (R): All polynomial functions are continuous everywhere on ℝ.
Answer: A — Both statements are true: f(0) = -1 and lim(x→0) (x³ + x² - 1) = -1, confirming continuity; and (R) correctly explains why all polynomials are continuous.
Q10. A function is defined as f(x) = {2x + 1 if x < 1, k if x = 1, x² if x > 1}. For what value of k is f continuous at x = 1?
Answer: C — For continuity at x = 1: lim(x→1−) (2x + 1) = 3 and lim(x→1+) x² = 1, but we need left limit = right limit = f(1); actually lim(x→1−) (2x + 1) = 3, so k = 3.
Definition of continuity of f at point c
Function f is continuous at c if lim(x→c) f(x) = f(c), meaning the left limit, right limit, and function value all exist and are equal.
When is a function called continuous on its domain?
A function is continuous if it is continuous at every point in its domain.
What does it mean graphically when a function is continuous?
The graph can be drawn without lifting the pen from the paper, showing no jumps, breaks, or holes at any point.
Example: Is f(x) = 2x + 3 continuous at x = 1?
Yes, because f(1) = 5 and lim(x→1) (2x + 3) = 5, so the limit equals the function value.
Example: Is f(x) = |x| continuous at x = 0?
Yes, because f(0) = 0, lim(x→0−) |x| = 0, and lim(x→0+) |x| = 0, so all three are equal.
What is a point of discontinuity?
A point c where the function is not continuous, meaning lim(x→c) f(x) ≠ f(c) or the limit does not exist.
Piecewise function: f(x) = 1 if x ≤ 0, and 2 if x > 0. Is it continuous at x = 0?
No, because lim(x→0−) f(x) = 1 but lim(x→0+) f(x) = 2, so left and right limits don't match.
For a function defined on [a,b], what condition ensures continuity at endpoint a?
The right-hand limit must exist and equal f(a), written as lim(x→a+) f(x) = f(a).
Is a constant function f(x) = k continuous everywhere?
Yes, because for any point c, f(c) = k and lim(x→c) k = k, satisfying the continuity condition.
Is the identity function f(x) = x continuous on all real numbers?
Yes, because for any point c, f(c) = c and lim(x→c) x = c, so continuity holds at every real number.
Define continuity of a function at a point c. Give one example of a function that is continuous at x = 0 and one that is discontinuous at x = 0. [2 marks]
State the three conditions: f(c) defined, limit exists, and limit = f(c). For continuous example use f(x) = |x| or x²; for discontinuous use a piecewise function with unequal one-sided limits at x = 0.
Examine the continuity of the function f(x) = {(x² - 9)/(x - 3) if x ≠ 3, and 6 if x = 3} at x = 3. Show all working steps and state whether it is continuous or discontinuous, and why. [5 marks]
Simplify (x² - 9)/(x - 3) to find lim(x→3) f(x); compare this limit with f(3) = 6; continuity requires all three values to match. Check: (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3 for x ≠ 3.
Prove that the function f(x) = |x| is continuous at every real number. Explain your proof with reference to the definition of continuity and handle the cases x < 0, x = 0, and x > 0 separately. [6 marks]
Rewrite |x| as a piecewise function: -x if x < 0, x if x ≥ 0. For each region and the boundary point x = 0, show lim(x→c) |x| = |c| = f(c) using polynomial continuity for x ≠ 0 and matching one-sided limits at x = 0. Conclude general continuity.
Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly
Try StudyOS Free →