Computing Limits for MATH 139
Exam Relevance for MATH 139
Tested on every exam. Master factoring, conjugates, and algebraic manipulation for indeterminate forms.
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What Does "Computing Limits" Mean?
Computing limits means finding the value that a function approaches as $x$ gets closer and closer to some number.
The Big Question: What happens to $f(x)$ as $x$ approaches $a$?
The Strategy
Step 1: Try Direct Substitution First!
Always start by plugging in the value:
- If you get a number → that's your answer!
- If you get $\frac{0}{0}$ → it's indeterminate, use algebraic tricks
- If you get $\frac{\text{number}}{0}$ → usually DNE or $\pm\infty$
Step 2: If Indeterminate ($\frac{0}{0}$), Try These Techniques
- Factor and cancel — most common method
- Rationalize — when you see square roots
- Expand — sometimes simplifies things
- Common denominator — for complex fractions
Technique 1: Factoring and Canceling
The most common technique! When both numerator and denominator equal zero, they share a common factor.
Problem: $\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$
Step 1: Try direct substitution: $$\frac{3^2 - 9}{3 - 3} = \frac{0}{0} \quad \text{Indeterminate!}$$
Step 2: Factor the numerator: $$x^2 - 9 = (x-3)(x+3)$$
Step 3: Cancel the common factor: $$\frac{(x-3)(x+3)}{x-3} = x + 3 \quad (x \neq 3)$$
Step 4: Now substitute: $$\lim_{x \to 3} (x + 3) = 3 + 3 = \boxed{6}$$
Problem: $\displaystyle\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$
Step 1: Check: $\frac{8 - 8}{0} = \frac{0}{0}$ ✓ Indeterminate
Step 2: Factor using the difference of cubes formula: $$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$ $$x^3 - 8 = x^3 - 2^3 = (x-2)(x^2 + 2x + 4)$$
Step 3: Cancel: $$\frac{(x-2)(x^2 + 2x + 4)}{x-2} = x^2 + 2x + 4$$
Step 4: Substitute: $$2^2 + 2(2) + 4 = 4 + 4 + 4 = \boxed{12}$$
Technique 2: Rationalizing
When you see square roots, multiply by the conjugate.
Conjugate pairs:
- $\sqrt{a} + b$ and $\sqrt{a} - b$
- $\sqrt{a} + \sqrt{b}$ and $\sqrt{a} - \sqrt{b}$
Why it works: $(\sqrt{a} + b)(\sqrt{a} - b) = a - b^2$ — the square root disappears!
Problem: $\displaystyle\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$
Step 1: Check: $\frac{\sqrt{4} - 2}{0} = \frac{0}{0}$ ✓ Indeterminate
Step 2: Multiply by the conjugate: $$\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2}$$
Step 3: Simplify the numerator: $$(\sqrt{x+4} - 2)(\sqrt{x+4} + 2) = (x+4) - 4 = x$$
Step 4: The expression becomes: $$\frac{x}{x(\sqrt{x+4} + 2)} = \frac{1}{\sqrt{x+4} + 2}$$
Step 5: Now substitute $x = 0$: $$\frac{1}{\sqrt{0+4} + 2} = \frac{1}{2 + 2} = \boxed{\frac{1}{4}}$$
Technique 3: Common Denominators
For complex fractions, combine into a single fraction first.
Problem: $\displaystyle\lim_{x \to 0} \frac{\frac{1}{x+2} - \frac{1}{2}}{x}$
Step 1: Check: plugging in gives $\frac{0}{0}$ ✓
Step 2: Combine the fractions in the numerator: $$\frac{1}{x+2} - \frac{1}{2} = \frac{2 - (x+2)}{2(x+2)} = \frac{-x}{2(x+2)}$$
Step 3: Rewrite the limit: $$\lim_{x \to 0} \frac{\frac{-x}{2(x+2)}}{x} = \lim_{x \to 0} \frac{-x}{2(x+2)} \cdot \frac{1}{x}$$
Step 4: Cancel the $x$: $$= \lim_{x \to 0} \frac{-1}{2(x+2)} = \frac{-1}{2(0+2)} = \boxed{-\frac{1}{4}}$$
Summary: The Flowchart
- Try plugging in the value
- If you get a number → done!
- If you get $\frac{0}{0}$:
- Look for factors to cancel
- If there's a square root, try rationalizing
- If there's a complex fraction, find common denominators
- Simplify and try plugging in again
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