Evaluating Limits Algebraically
Apply limit laws (sum, product, quotient, power) and evaluate limits algebraically by direct substitution, factoring and canceling, rationalizing, and using the squeeze theorem conceptually.
A lesson of Pre-Calculus, in Limits: An Introduction to Calculus, about 10 minutes.
By the end of this lesson, you will be able to evaluate limits using the limit laws, direct substitution, factoring, rationalization, and the Squeeze Theorem.
Limit Laws
If $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$:
| Law | Formula |
|---|---|
| Sum/Difference | $\lim[f \pm g] = L \pm M$ |
| Product | $\lim[f \cdot g] = L \cdot M$ |
| Quotient | $\lim[f/g] = L/M$ (if $M \neq 0$) |
| Scalar | $\lim[kf] = kL$ |
| Power | $\lim[f^n] = L^n$ |
| Root | $\lim\sqrt[n]{f} = \sqrt[n]{L}$ (appropriate conditions) |
Direct Substitution: if $f$ is a polynomial or rational function (and denominator ≠ 0 at $c$):
$$\lim_{x \to c} f(x) = f(c)$$
Indeterminate Forms: $\frac{0}{0}$
When substitution gives $0/0$, the limit may still exist. Use algebraic simplification.
Common techniques:
- Factor and cancel
- Rationalize (for square roots)
- Expand and simplify
Example 1 — Direct Substitution
$$\lim_{x \to 3}(2x^2 - 5x + 1) = 2(9) - 15 + 1 = 4$$
Example 2 — Factor and Cancel
$$\lim_{x \to 4}\frac{x^2 - 16}{x - 4} = \lim_{x \to 4}\frac{(x-4)(x+4)}{x-4} = \lim_{x \to 4}(x+4) = 8$$
Example 3 — Rationalization
$$\lim_{x \to 0}\frac{\sqrt{x+9} - 3}{x}$$
Multiply by $\dfrac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$:
$$= \lim_{x \to 0}\frac{(x+9) - 9}{x(\sqrt{x+9}+3)} = \lim_{x \to 0}\frac{x}{x(\sqrt{x+9}+3)} = \lim_{x \to 0}\frac{1}{\sqrt{x+9}+3} = \frac{1}{6}$$
Example 4 — Trigonometric Limit
Two important trig limits (which calculus proves):
$$\lim_{x \to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x \to 0}\frac{1-\cos x}{x} = 0$$
Using the first:
$$\lim_{x \to 0}\frac{\sin 3x}{x} = \lim_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} = 3(1) = 3$$