Unit 1.11 – Defining Continuity at a Point

How to Prove a Function is Continuous at \(x = c\):

1. STEP 1: Check if \(f(c)\) exists
   • Substitute \(x = c\) into the function
   • If you get a real number, condition 1 is satisfied ✓
   • If undefined, stop—discontinuous at \(c\) ✗
2. STEP 2: Check if \(\lim_{x \to c} f(x)\) exists
   • Find \(\lim_{x \to c^-} f(x)\) (left-hand limit)
   • Find \(\lim_{x \to c^+} f(x)\) (right-hand limit)
   • If they're equal and finite, condition 2 is satisfied ✓
   • If different or infinite, stop—discontinuous at \(c\) ✗
3. STEP 3: Check if \(\lim_{x \to c} f(x) = f(c)\)
   • Compare the limit (from Step 2) with the function value (from Step 1)
   • If they're equal, condition 3 is satisfied ✓
   • If different, discontinuous at \(c\) ✗
4. CONCLUSION: If all three conditions pass, the function is continuous at \(x = c\)! 🎉

Function: \(f(x) = 2x^2 + 3x - 1\) at \(x = 2\)

Solution:

1. Condition 1: \(f(2) = 2(2)^2 + 3(2) - 1 = 8 + 6 - 1 = 13\) ✓ (exists)
2. Condition 2: \(\lim_{x \to 2} (2x^2 + 3x - 1) = 13\) ✓ (polynomials are continuous)
3. Condition 3: \(\lim_{x \to 2} f(x) = 13 = f(2)\) ✓ (limit equals function value)

Conclusion: \(f(x)\) is continuous at \(x = 2\)

Function: \(g(x) = \frac{x^2 - 4}{x - 2}\) at \(x = 2\)

Solution:

1. Condition 1: \(g(2) = \frac{0}{0}\) — undefined ✗ (fails!)
2. Condition 2: Factor: \(g(x) = \frac{(x-2)(x+2)}{x-2} = x + 2\) for \(x \neq 2\)
   \(\lim_{x \to 2} g(x) = \lim_{x \to 2} (x+2) = 4\) ✓ (exists)
3. Condition 3: \(\lim_{x \to 2} g(x) = 4\), but \(g(2)\) undefined, so can't equal ✗ (fails!)

Conclusion: \(g(x)\) is NOT continuous at \(x = 2\) (removable discontinuity—hole at (2, 4))

Function: \(h(x) = \begin{cases} x^2 & x < 2 \\ 4 & x = 2 \\ x + 2 & x > 2 \end{cases}\) at \(x = 2\)

Solution:

1. Condition 1: \(h(2) = 4\) ✓ (defined)
2. Condition 2: Check one-sided limits:
   • \(\lim_{x \to 2^-} h(x) = \lim_{x \to 2^-} x^2 = 4\)
   • \(\lim_{x \to 2^+} h(x) = \lim_{x \to 2^+} (x+2) = 4\)
   • Since both equal 4, \(\lim_{x \to 2} h(x) = 4\) ✓
3. Condition 3: \(\lim_{x \to 2} h(x) = 4 = h(2)\) ✓

Conclusion: \(h(x)\) is continuous at \(x = 2\)

Function: \(p(x) = \begin{cases} x - 1 & x < 1 \\ 2 - x & x \geq 1 \end{cases}\) at \(x = 1\)

Solution:

1. Condition 1: \(p(1) = 2 - 1 = 1\) ✓ (defined)
2. Condition 2: Check one-sided limits:
   • \(\lim_{x \to 1^-} p(x) = \lim_{x \to 1^-} (x-1) = 0\)
   • \(\lim_{x \to 1^+} p(x) = \lim_{x \to 1^+} (2-x) = 1\)
   • Since \(0 \neq 1\), \(\lim_{x \to 1} p(x)\) does NOT exist ✗

Conclusion: \(p(x)\) is NOT continuous at \(x = 1\) (jump discontinuity)

Function: \(q(x) = \frac{1}{x - 3}\) at \(x = 3\)

Solution:

1. Condition 1: \(q(3) = \frac{1}{0}\) — undefined ✗
2. Condition 2: \(\lim_{x \to 3} \frac{1}{x-3} = \pm\infty\) — does NOT exist (infinite) ✗

Conclusion: \(q(x)\) is NOT continuous at \(x = 3\) (infinite discontinuity—vertical asymptote)

Determine if each function is continuous at the given point. Justify using all three conditions.

1. \(f(x) = 3x + 5\) at \(x = 2\)
2. \(g(x) = \frac{x^2 - 9}{x - 3}\) at \(x = 3\)
3. \(h(x) = \begin{cases} x^2 & x \leq 1 \\ 2x - 1 & x > 1 \end{cases}\) at \(x = 1\)
4. \(p(x) = \frac{2}{x - 4}\) at \(x = 4\)

Answers:

1. Continuous — All three conditions satisfied (polynomial)
2. NOT continuous — \(g(3)\) undefined (removable, hole at (3, 6))
3. Continuous — Left limit = 1, right limit = 1, \(h(1) = 1\) ✓
4. NOT continuous — Infinite discontinuity (vertical asymptote at x = 4)