Lesson 2: Computing Limits Algebraically
Estimated time: 35-45 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Apply direct substitution to evaluate limits of continuous functions
- State and use the basic limit laws (sum, product, quotient, power)
- Evaluate limits by factoring and canceling common factors
- Use rationalization (conjugate multiplication) to compute limits involving radicals
- Apply the Squeeze Theorem to evaluate limits
Direct Substitution
The simplest strategy for evaluating a limit is direct substitution: just plug in the target value. This works whenever the function is continuous at the target.
Direct Substitution Property
If f is a polynomial or rational function and a is in the domain of f, then limx→a f(x) = f(a).
Example 1: Polynomial Limit
Evaluate limx→3 (2x² − 5x + 1).
Solution: Since polynomials are continuous everywhere, substitute directly:
= 2(3)² − 5(3) + 1 = 18 − 15 + 1 = 4
Example 2: Rational Function Limit
Evaluate limx→2 (x + 1)/(x² + 3).
Solution: Check the denominator: 2² + 3 = 7 ≠ 0, so substitute:
= (2 + 1)/(4 + 3) = 3/7
Limit Laws
When direct substitution does not work immediately, the limit laws let us break complicated limits into simpler pieces.
Basic Limit Laws
Suppose limx→a f(x) = L and limx→a g(x) = M. Then:
- Sum Law: limx→a [f(x) + g(x)] = L + M
- Difference Law: limx→a [f(x) − g(x)] = L − M
- Constant Multiple Law: limx→a [c · f(x)] = c · L
- Product Law: limx→a [f(x) · g(x)] = L · M
- Quotient Law: limx→a [f(x)/g(x)] = L/M, provided M ≠ 0
- Power Law: limx→a [f(x)]n = Ln
- Root Law: limx→a √[f(x)] = √L (when L ≥ 0 for even roots)
Example 3: Using Limit Laws
Given limx→1 f(x) = 4 and limx→1 g(x) = −2, find limx→1 [3f(x) − g(x)²].
Step 1: By the constant multiple law: lim 3f(x) = 3(4) = 12
Step 2: By the power law: lim [g(x)]² = (−2)² = 4
Step 3: By the difference law: 12 − 4 = 8
Limits by Factoring
When direct substitution gives the indeterminate form 0/0, try factoring the numerator and denominator to cancel the common factor causing the zero.
Example 4: Factor and Cancel
Evaluate limx→3 (x² − 9)/(x − 3).
Step 1: Direct substitution gives 0/0 (indeterminate).
Step 2: Factor the numerator: x² − 9 = (x − 3)(x + 3).
Step 3: Cancel: (x − 3)(x + 3)/(x − 3) = x + 3 for x ≠ 3.
Step 4: Now substitute: limx→3 (x + 3) = 3 + 3 = 6.
Example 5: Factoring a Cubic
Evaluate limx→−1 (x³ + 1)/(x + 1).
Step 1: Direct substitution: 0/0, indeterminate.
Step 2: Factor sum of cubes: x³ + 1 = (x + 1)(x² − x + 1).
Step 3: Cancel: (x + 1)(x² − x + 1)/(x + 1) = x² − x + 1.
Step 4: Substitute: (−1)² − (−1) + 1 = 1 + 1 + 1 = 3.
Limits by Rationalization
When a limit involves a square root and gives 0/0, multiply numerator and denominator by the conjugate to eliminate the radical.
Conjugate
The conjugate of √a + b is √a − b, and vice versa. Multiplying conjugates uses the difference of squares: (√a + b)(√a − b) = a − b².
Example 6: Rationalization
Evaluate limx→0 (√(x + 4) − 2)/x.
Step 1: Substitution gives (√4 − 2)/0 = 0/0.
Step 2: Multiply by the conjugate:
[(√(x+4) − 2)/x] · [(√(x+4) + 2)/(√(x+4) + 2)]
= [(x + 4) − 4] / [x(√(x+4) + 2)]
= x / [x(√(x+4) + 2)]
Step 3: Cancel x: = 1/(√(x+4) + 2)
Step 4: Substitute x = 0: = 1/(√4 + 2) = 1/(2 + 2) = 1/4.
The Squeeze Theorem
The Squeeze Theorem (also called the Sandwich Theorem) lets us find limits by trapping a function between two others whose limits we know.
Squeeze Theorem
If g(x) ≤ f(x) ≤ h(x) for all x near a (except possibly at a), and limx→a g(x) = limx→a h(x) = L, then limx→a f(x) = L.
Example 7: Using the Squeeze Theorem
Show that limx→0 x² sin(1/x) = 0.
Step 1: We know −1 ≤ sin(1/x) ≤ 1 for all x ≠ 0.
Step 2: Multiply by x² (which is non-negative): −x² ≤ x² sin(1/x) ≤ x².
Step 3: limx→0 (−x²) = 0 and limx→0 x² = 0.
Step 4: By the Squeeze Theorem: limx→0 x² sin(1/x) = 0.
Special Trigonometric Limits
Two important limits involving trigonometric functions appear frequently. These should be memorized.
Fundamental Trig Limits
1. limx→0 (sin x)/x = 1
2. limx→0 (1 − cos x)/x = 0
Example 8: Using the Fundamental Trig Limit
Evaluate limx→0 (sin 5x)/(3x).
Step 1: Rewrite to match the pattern (sin u)/u:
(sin 5x)/(3x) = (5/3) · (sin 5x)/(5x)
Step 2: Let u = 5x. As x → 0, u → 0:
= (5/3) · limu→0 (sin u)/u = (5/3)(1) = 5/3.
Strategy Summary
When evaluating limx→a f(x), follow this strategy:
- Try direct substitution. If you get a number, that is the answer.
- If you get 0/0, try algebraic simplification: factor, rationalize, or simplify complex fractions.
- If you get nonzero/0, the limit is ±∞ or DNE — check sign from each side.
- If trig functions are involved, try rewriting using (sin x)/x or (1 − cos x)/x.
- If trapped, try the Squeeze Theorem.
Key Takeaways
- Direct substitution works for polynomials, rational functions (when denominator is nonzero), trig, exponential, and log functions at points in their domain.
- The limit laws let you break a complex limit into simpler pieces (sum, product, quotient, power).
- An indeterminate form 0/0 signals that algebraic simplification is needed (factor, rationalize, or rewrite).
- Rationalization multiplies by the conjugate to clear radicals.
- The Squeeze Theorem bounds a function between two simpler ones.
- The special trig limits lim (sin x)/x = 1 and lim (1 − cos x)/x = 0 are essential tools.
Check Your Understanding
1. Evaluate limx→4 (x² − 16)/(x − 4).
2. Evaluate limx→0 (√(x+9) − 3)/x.
3. Evaluate limx→0 (sin 3x)/(sin 5x).
4. What does the Squeeze Theorem require?
Ready for More?
Next Lesson
In Lesson 3 you will explore limits at infinity, end behavior, and asymptotes.
Start Lesson 3