Limits and Continuity - Main Theorem

The limit laws provide the algebraic foundation for computing limits without resorting to the epsilon-delta definition each time. These rules allow us to break down complex limits into simpler components.

TheoremLimit Laws

Suppose $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ both exist. Then:

Sum Rule: $\lim_{x \to a} [f(x) + g(x)] = L + M$
Difference Rule: $\lim_{x \to a} [f(x) - g(x)] = L - M$
Constant Multiple Rule: $\lim_{x \to a} [cf(x)] = cL$ for any constant $c$
Product Rule: $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
Quotient Rule: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ provided $M \neq 0$
Power Rule: $\lim_{x \to a} [f(x)]^n = L^n$ for any positive integer $n$
Root Rule: $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ provided $L > 0$ when $n$ is even

These laws are intuitive but require rigorous proof using the epsilon-delta definition. They extend to one-sided limits and limits at infinity with appropriate modifications.

ExampleUsing Limit Laws

Compute $\lim_{x \to 3} \frac{x^2 + 2x - 15}{x - 3}$ .

Direct substitution gives $\frac{0}{0}$ , an indeterminate form. Factor the numerator: $x^2 + 2x - 15 = (x - 3)(x + 5)$

For $x \neq 3$ : $\frac{x^2 + 2x - 15}{x - 3} = \frac{(x - 3)(x + 5)}{x - 3} = x + 5$

By the limit laws: $\lim_{x \to 3} \frac{x^2 + 2x - 15}{x - 3} = \lim_{x \to 3} (x + 5) = 3 + 5 = 8$

TheoremSqueeze Theorem

Suppose $g(x) \leq f(x) \leq h(x)$ for all $x$ in an open interval containing $a$ , except possibly at $a$ itself. If $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ then $\lim_{x \to a} f(x) = L$

The Squeeze Theorem is particularly powerful for functions that oscillate or are difficult to evaluate directly.

ExampleA Classic Application

Prove that $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$ .

We know $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$ . Multiplying by $x^2 \geq 0$ : $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$

Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$ , by the Squeeze Theorem: $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

Note that $\sin\left(\frac{1}{x}\right)$ oscillates infinitely as $x \to 0$ and has no limit, but multiplication by $x^2$ dampens these oscillations.

Remark

The Squeeze Theorem is often the only practical way to evaluate limits involving trigonometric functions with oscillatory behavior or products where one factor is bounded and another approaches zero.