Grade: Grade 12 Subject: Mathematics Unit: Precalculus Completion SAT: AdvancedMath ACT: Math

Introduction to Limits

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What is a Limit?

Limits are the foundation of calculus. They describe what value a function approaches as the input approaches some value. Understanding limits helps us analyze function behavior at points where direct substitution might not work.

Limit Definition

The limit of f(x) as x approaches a is L, written:

lim (x→a) f(x) = L

This means that as x gets arbitrarily close to a (but not equal to a), f(x) gets arbitrarily close to L.

Key Insight

A limit describes the behavior of a function near a point, not necessarily the value at that point. The function doesn't even need to be defined at a for the limit to exist!

Evaluating Limits: Direct Substitution

For many functions, you can find the limit by simply substituting the value:

Direct Substitution

If f(x) is continuous at x = a, then:

lim (x→a) f(x) = f(a)

This works for polynomials, rational functions (when denominator ≠ 0), exponentials, and trigonometric functions at most points.

Indeterminate Forms

Sometimes direct substitution gives meaningless results like 0/0 or ∞/∞. These are called indeterminate forms and require additional techniques.

Indeterminate FormWhat It Looks LikeStrategy
0/0Both numerator and denominator → 0Factor, simplify, or use L'Hopital's rule
∞/∞Both numerator and denominator → ∞Divide by highest power, L'Hopital's rule
0 · ∞One factor → 0, other → ∞Rewrite as fraction
∞ - ∞Subtraction of infinitiesCombine into single fraction

Techniques for Evaluating Limits

1. Factoring and Canceling

When you get 0/0, factor the numerator and denominator, then cancel common factors.

Example: lim (x→2) (x² - 4)/(x - 2) = lim (x→2) (x+2)(x-2)/(x-2) = lim (x→2) (x+2) = 4

2. Rationalizing

When expressions contain square roots, multiply by the conjugate.

Example: For (√x - 2)/(x - 4), multiply by (√x + 2)/(√x + 2)

3. Limits at Infinity

To find lim (x→∞) of a rational function, divide all terms by the highest power of x in the denominator.

  • If degree of numerator < degree of denominator: limit = 0
  • If degrees are equal: limit = ratio of leading coefficients
  • If degree of numerator > degree of denominator: limit = ±∞

One-Sided Limits

Left and Right Limits

Left-hand limit: lim (x→a⁻) f(x) - approaching from values less than a

Right-hand limit: lim (x→a⁺) f(x) - approaching from values greater than a

The two-sided limit exists if and only if both one-sided limits exist and are equal.

Limit Laws

Properties of Limits

If lim (x→a) f(x) = L and lim (x→a) g(x) = M, then:

  • Sum: lim [f(x) + g(x)] = L + M
  • Difference: lim [f(x) - g(x)] = L - M
  • Product: lim [f(x) · g(x)] = L · M
  • Quotient: lim [f(x)/g(x)] = L/M (if M ≠ 0)
  • Constant Multiple: lim [c · f(x)] = c · L
  • Power: lim [f(x)]ⁿ = Lⁿ

Special Limits to Know

LimitValueApplication
lim (x→0) (sin x)/x1Trigonometric limits
lim (x→0) (1 - cos x)/x0Trigonometric limits
lim (x→∞) (1 + 1/x)^xe ≈ 2.718Natural exponential
lim (x→0) (eˣ - 1)/x1Exponential limits

Continuity and Limits

Continuous Function

A function f is continuous at x = a if:

  1. f(a) is defined
  2. lim (x→a) f(x) exists
  3. lim (x→a) f(x) = f(a)

If any condition fails, there is a discontinuity at x = a.

Examples

Example 1: Direct Substitution

Problem: Find lim (x→3) (2x² - 5x + 1)

Solution:

This is a polynomial, so direct substitution works:

lim (x→3) (2x² - 5x + 1) = 2(3)² - 5(3) + 1

= 2(9) - 15 + 1

= 18 - 15 + 1

= 4

Example 2: Factoring to Resolve 0/0

Problem: Find lim (x→-3) (x² + 5x + 6)/(x + 3)

Solution:

Step 1: Check direct substitution: (-3)² + 5(-3) + 6 / (-3 + 3) = 0/0 (indeterminate)

Step 2: Factor the numerator

x² + 5x + 6 = (x + 2)(x + 3)

Step 3: Cancel common factors

lim (x→-3) (x + 2)(x + 3)/(x + 3) = lim (x→-3) (x + 2)

Step 4: Substitute

= -3 + 2 = -1

Example 3: Rationalizing

Problem: Find lim (x→4) (√x - 2)/(x - 4)

Solution:

Step 1: Direct substitution gives (2-2)/(4-4) = 0/0

Step 2: Multiply by conjugate

= lim (x→4) [(√x - 2)/(x - 4)] · [(√x + 2)/(√x + 2)]

= lim (x→4) (x - 4)/[(x - 4)(√x + 2)]

Step 3: Cancel (x - 4)

= lim (x→4) 1/(√x + 2)

Step 4: Substitute

= 1/(√4 + 2) = 1/(2 + 2) = 1/4

Example 4: Limit at Infinity

Problem: Find lim (x→∞) (3x² - 2x + 1)/(5x² + 4)

Solution:

Step 1: Both numerator and denominator → ∞

Step 2: Divide all terms by x² (highest power in denominator)

= lim (x→∞) (3 - 2/x + 1/x²)/(5 + 4/x²)

Step 3: As x → ∞, terms with x in denominator → 0

= (3 - 0 + 0)/(5 + 0)

= 3/5

Note: Since degrees are equal, the limit is the ratio of leading coefficients.

Example 5: One-Sided Limits

Problem: Find the left and right limits of f(x) = |x|/x at x = 0

Solution:

Right-hand limit (x → 0⁺):

When x > 0: |x| = x, so |x|/x = x/x = 1

lim (x→0⁺) |x|/x = 1

Left-hand limit (x → 0⁻):

When x < 0: |x| = -x, so |x|/x = -x/x = -1

lim (x→0⁻) |x|/x = -1

Conclusion: Since 1 ≠ -1, the two-sided limit lim (x→0) |x|/x does not exist.

Practice

Evaluate each limit using appropriate techniques.

1. lim (x→2) (x² + 3x - 2) = ?

A) 6   B) 8   C) 10   D) 12

2. lim (x→1) (x² - 1)/(x - 1) = ?

A) 0   B) 1   C) 2   D) Does not exist

3. lim (x→∞) (4x - 3)/(2x + 5) = ?

A) 0   B) 2   C) 4   D) ∞

4. lim (x→0) sin(3x)/x = ?

A) 0   B) 1   C) 3   D) Does not exist

5. lim (x→9) (√x - 3)/(x - 9) = ?

A) 1/6   B) 1/3   C) 3   D) 6

6. lim (x→∞) 5x/(x² + 1) = ?

A) 0   B) 1   C) 5   D) ∞

7. If lim (x→a) f(x) = 4 and lim (x→a) g(x) = -2, find lim (x→a) [3f(x) - 2g(x)].

A) 8   B) 10   C) 14   D) 16

8. lim (x→-2) (x³ + 8)/(x + 2) = ?

A) 4   B) 8   C) 12   D) 16

9. For what value of c is f(x) = {x² if x ≤ 2; cx if x > 2} continuous at x = 2?

A) 1   B) 2   C) 4   D) 8

10. lim (x→0⁺) 1/x = ?

A) 0   B) 1   C) -∞   D) +∞

Click to reveal answers
  1. B) 8 - Direct substitution: 4 + 6 - 2 = 8
  2. C) 2 - Factor: (x+1)(x-1)/(x-1) = x+1, then substitute x=1
  3. B) 2 - Ratio of leading coefficients: 4/2 = 2
  4. C) 3 - Use sin(u)/u → 1 as u→0, here u=3x, so limit = 3(1) = 3
  5. A) 1/6 - Rationalize: 1/(√x+3) at x=9 gives 1/6
  6. A) 0 - Degree of numerator < degree of denominator
  7. D) 16 - 3(4) - 2(-2) = 12 + 4 = 16
  8. C) 12 - Factor x³+8 = (x+2)(x²-2x+4), cancel, substitute: 4+4+4=12
  9. B) 2 - Need 2² = c(2), so 4 = 2c, c = 2
  10. D) +∞ - As x approaches 0 from the right, 1/x grows without bound

Check Your Understanding

1. Explain why lim (x→2) f(x) might exist even if f(2) is undefined. Give an example.

Show answer

A limit describes what value f(x) approaches as x gets close to 2, regardless of what happens exactly at x = 2. Example: f(x) = (x² - 4)/(x - 2). This function is undefined at x = 2 (division by zero), but lim (x→2) f(x) = 4 because as x approaches 2, the expression approaches 4. The limit exists because we only care about behavior near the point, not at the point itself.

2. Why can't we just "plug in" to find lim (x→0) (sin x)/x? What's special about this limit?

Show answer

Plugging in gives sin(0)/0 = 0/0, which is indeterminate. This limit is special because it equals 1, and this result is fundamental in calculus - it's used to derive the derivatives of trigonometric functions. The proof requires the squeeze theorem or geometric arguments showing that as x → 0, sin x behaves almost exactly like x.

3. What does it mean for a function to be continuous, and why do discontinuities matter?

Show answer

A function is continuous at a point if you can draw it without lifting your pencil - formally, the limit equals the function value. Discontinuities matter because: (1) calculus tools like derivatives require continuity, (2) real-world models often assume continuity (continuous change), (3) discontinuities indicate special behavior like jumps, holes, or asymptotes that need separate analysis, (4) many theorems (like Intermediate Value Theorem) only apply to continuous functions.

4. Compare and contrast the strategies for finding limits at a finite point versus limits at infinity.

Show answer

At a finite point: Try direct substitution first. If you get 0/0 or similar indeterminate forms, use factoring, rationalizing, or algebraic manipulation. Check one-sided limits if needed. At infinity: Direct substitution doesn't make sense. Instead, identify dominant terms (highest powers), divide all terms by the highest power in the denominator, and observe which terms approach zero. Compare degrees: if numerator degree is less, limit is 0; if equal, it's the ratio of leading coefficients; if greater, limit is infinite.

🚀 Next Steps

  • Review any concepts that felt challenging
  • Move on to the next lesson when ready
  • Return to practice problems periodically for review