Limit Laws and Algebraic Evaluation
State the sum, difference, product, quotient, and power limit laws. Evaluate limits by direct substitution, factoring and canceling, rationalizing numerators/denominators, and applying the squeeze theorem.
A lesson of Calculus, in Limits and Continuity, about 10 minutes.
By the end of this lesson, you will be able to state and apply all limit laws, evaluate limits by direct substitution, factoring, rationalizing, and simplifying complex fractions, and resolve all forms of indeterminate expressions at this level.
The Limit Laws
If $\lim_{x \to c}f(x) = L$ and $\lim_{x \to c}g(x) = M$, then:
- Sum: $\lim[f \pm g] = L \pm M$
- Product: $\lim[fg] = LM$
- Quotient: $\lim[f/g] = L/M$ if $M \neq 0$
- Scalar: $\lim[kf] = kL$
- Power: $\lim[f^n] = L^n$ ($n$ positive integer)
- Root: $\lim\sqrt[n]{f} = \sqrt[n]{L}$ (when defined)
- Substitution: if $f$ is a polynomial or rational (denominator $\neq 0$ at $c$): $\lim_{x\to c}f(x) = f(c)$
Indeterminate Forms at This Level
$\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$
These require algebraic manipulation to resolve. $L$'Hôpital's Rule (covered later) handles many systematically.
Technique 1: Factor and Cancel
$\lim_{x \to -3}\dfrac{x^2+5x+6}{x^2-x-12}$
$= \lim_{x \to -3}\dfrac{(x+2)(x+3)}{(x-4)(x+3)} = \lim_{x \to -3}\dfrac{x+2}{x-4} = \dfrac{-1}{-7} = \dfrac{1}{7}$
Technique 2: Rationalize
$\lim_{x \to 1}\dfrac{\sqrt{x}-1}{x-1}$
Multiply by $\dfrac{\sqrt{x}+1}{\sqrt{x}+1}$:
$= \lim_{x \to 1}\dfrac{x-1}{(x-1)(\sqrt{x}+1)} = \lim_{x \to 1}\dfrac{1}{\sqrt{x}+1} = \dfrac{1}{2}$
Technique 3: Simplify Complex Fractions
$\lim_{x \to 0}\dfrac{\frac{1}{x+2} - \frac{1}{2}}{x}$
$= \lim_{x \to 0}\dfrac{\frac{2-(x+2)}{2(x+2)}}{x} = \lim_{x \to 0}\dfrac{-x}{2x(x+2)} = \lim_{x \to 0}\dfrac{-1}{2(x+2)} = -\dfrac{1}{4}$
Technique 4: Special Limits
$\lim_{x \to 0}\dfrac{\sin(3x)}{5x}$
$= \lim_{x \to 0}\dfrac{3}{5}\cdot\dfrac{\sin(3x)}{3x} = \dfrac{3}{5}\cdot 1 = \dfrac{3}{5}$
$\lim_{x \to 0}\dfrac{\tan(4x)}{\sin(6x)}$
$= \lim_{x \to 0}\dfrac{\sin(4x)}{\cos(4x)\cdot\sin(6x)} = \lim_{x \to 0}\dfrac{\sin(4x)}{4x}\cdot\dfrac{6x}{\sin(6x)}\cdot\dfrac{4x}{6x\cos(4x)}$
$= 1\cdot 1\cdot\dfrac{4}{6} = \dfrac{2}{3}$
Example: $\frac{\infty}{\infty}$ Form
$\lim_{x \to \infty}\dfrac{3x^2-5}{2x^2+7}$
Divide by $x^2$: $= \lim_{x \to \infty}\dfrac{3 - 5/x^2}{2 + 7/x^2} = \dfrac{3}{2}$