What is Continuity?

Understanding when a function is continuous at a point

Understanding Continuity

A function is continuous if you can draw it without lifting your pencil. Mathematically, it's more precise!

The Informal Definition

A function is continuous at x = a if:

The graph has no breaks, jumps, or holes at that point

The Formal Definition

A function f is continuous at x = a if all three conditions hold:

f(a) exists - The function is defined at a
$\lim_{x \to a} f(x)$ exists - The limit exists at a
$\lim_{x \to a} f(x) = f(a)$ - The limit equals the function value

All three must be true! If even one fails, the function is discontinuous at that point.

Testing for Continuity

Example 1: Is $f(x) = x^2$ continuous at x = 3?

Check 1: Does f(3) exist? $f(3) = 3^2 = 9$ ✓ Yes

Check 2: Does $\lim_{x \to 3} f(x)$ exist? $\lim_{x \to 3} x^2 = 9$ ✓ Yes

Check 3: Does $\lim_{x \to 3} f(x) = f(3)$ ? $9 = 9$ ✓ Yes

Conclusion: f is continuous at x = 3!

Example 2: Not Continuous

$f(x) = \begin{cases} x + 1 & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases}$

Is this continuous at x = 2?

Check 1: Does f(2) exist? $f(2) = 5$ ✓ Yes

Check 2: Does $\lim_{x \to 2} f(x)$ exist? $\lim_{x \to 2} (x + 1) = 3$ ✓ Yes

Check 3: Does $\lim_{x \to 2} f(x) = f(2)$ ? $3 \neq 5$ ✗ NO!

Conclusion: f is NOT continuous at x = 2 (even though f(2) exists and the limit exists!)

Continuous Everywhere

Many familiar functions are continuous everywhere:

Polynomials: $x^2$ , $x^3 - 2x + 1$ , etc.
Exponential functions: $e^x$ , $2^x$
Sine and cosine: $\sin(x)$ , $\cos(x)$
Square root (on its domain): $\sqrt{x}$ for $x \geq 0$

Common Discontinuities

Functions are typically discontinuous where:

Rational functions have division by zero
Piecewise functions change formulas
Absolute value creates a corner (still continuous, but not differentiable!)

Why Continuity Matters

Continuous functions have nice properties:

Intermediate Value Theorem: If continuous on [a, b], it takes every value between f(a) and f(b)
Can find limits by direct substitution
Behave predictably - no surprises!

Visual Test

On a graph, a function is continuous at a point if:

No gap (both sides connect)
No hole (no open circle)
No jump (no sudden leap)
The point is actually on the curve

📚 Practice Problems

1Problem 1medium

❓ Question:

Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at x = 3.

💡 Show Solution

Check 1: Does f(3) exist?

$f(3) = \frac{3^2 - 9}{3 - 3} = \frac{0}{0}$

This is undefined! ✗

Conclusion: f is NOT continuous at x = 3 because f(3) doesn't exist.

The function has a hole at x = 3.

Note: Even though we can find the limit: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x - 3} = \lim_{x \to 3} (x + 3) = 6$

The limit exists (equals 6), but f(3) doesn't exist, so the function fails condition 1 of continuity.

2Problem 2hard

❓ Question:

For what value of k is $f(x) = \begin{cases} x^2 & \text{if } x < 2 \\ kx + 1 & \text{if } x \geq 2 \end{cases}$ continuous at x = 2?

💡 Show Solution

For f to be continuous at x = 2, we need: $\lim_{x \to 2} f(x) = f(2)$

Find f(2): Since $x = 2 \geq 2$ , we use the second piece: $f(2) = k(2) + 1 = 2k + 1$

Find the limit from the left: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 4$

Find the limit from the right: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (kx + 1) = 2k + 1$

For continuity, these must all be equal: $4 = 2k + 1$ $3 = 2k$ $k = \frac{3}{2}$

Answer: $k = \frac{3}{2}$

With this value, f(2) = 4 and both one-sided limits equal 4, making the function continuous!

3Problem 3easy

❓ Question:

Is f(x) = x² - 3x + 5 continuous at x = 2?

💡 Show Solution

Step 1: Check the three conditions for continuity at x = a:

f(a) is defined
lim(x→a) f(x) exists
lim(x→a) f(x) = f(a)

Step 2: Check if f(2) is defined: f(2) = (2)² - 3(2) + 5 = 4 - 6 + 5 = 3 ✓ f(2) is defined

Step 3: Find lim(x→2) f(x): Since this is a polynomial, use direct substitution lim(x→2) (x² - 3x + 5) = 3 ✓ Limit exists

Step 4: Compare: lim(x→2) f(x) = 3 = f(2) ✓ All three conditions satisfied

Answer: Yes, f(x) is continuous at x = 2

4Problem 4medium

❓ Question:

Determine if f(x) = { x + 1, if x < 3; 5, if x = 3; 2x - 1, if x > 3 } is continuous at x = 3.

💡 Show Solution

Step 1: Check if f(3) is defined: f(3) = 5 ✓

Step 2: Find lim(x→3⁻) f(x): For x < 3, use f(x) = x + 1 lim(x→3⁻) (x + 1) = 3 + 1 = 4

Step 3: Find lim(x→3⁺) f(x): For x > 3, use f(x) = 2x - 1 lim(x→3⁺) (2x - 1) = 2(3) - 1 = 5

Step 4: Check if two-sided limit exists: lim(x→3⁻) f(x) = 4 lim(x→3⁺) f(x) = 5 Since 4 ≠ 5, lim(x→3) f(x) does not exist ✗

Step 5: Conclusion: Since the limit doesn't exist, f is not continuous at x = 3

Answer: No, f(x) is not continuous at x = 3

5Problem 5hard

❓ Question:

Find the value of k that makes f(x) = { (x² - 4)/(x - 2), if x ≠ 2; k, if x = 2 } continuous at x = 2.

💡 Show Solution

Step 1: For continuity at x = 2, need: lim(x→2) f(x) = f(2) = k

Step 2: Find lim(x→2) (x² - 4)/(x - 2): Direct substitution gives 0/0 (indeterminate)

Step 3: Factor numerator: x² - 4 = (x - 2)(x + 2)

Step 4: Simplify: (x² - 4)/(x - 2) = [(x - 2)(x + 2)]/(x - 2) = x + 2 (for x ≠ 2)

Step 5: Evaluate limit: lim(x→2) (x + 2) = 2 + 2 = 4

Step 6: Set equal to f(2): k = 4

Step 7: Verify: With k = 4, lim(x→2) f(x) = 4 = f(2) ✓

Answer: k = 4

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