Limits & Continuity

Definition of a Limit

$\lim_{x \to c} f(x) = L$ means $f(x)$ approaches $L$ as $x$ approaches $c$.

Techniques for Evaluating Limits

1. Direct substitution: try plugging in $c$ first.
2. Factoring: cancel common factors.
3. Rationalizing: multiply by the conjugate.
4. L'Hôpital's Rule: if $0/0$ or $\infty/\infty$, then $\lim f/g = \lim f'/g'$.

One-Sided Limits

$\lim_{x \to c^+} f(x) \quad \text{and} \quad \lim_{x \to c^-} f(x)$ The two-sided limit exists only if both one-sided limits exist and are equal.

Limits at Infinity

$\lim_{x \to \infty} \frac{P(x)}{Q(x)}$

• Same degree: ratio of leading coefficients.
• Higher degree in numerator: $\pm\infty$.
• Higher degree in denominator: $0$.

Continuity

A function $f$ is continuous at $x = c$ if:

1. $f(c)$ is defined.
2. $\lim_{x \to c} f(x)$ exists.
3. $\lim_{x \to c} f(x) = f(c)$.

Important Theorems

• Intermediate Value Theorem (IVT): If $f$ is continuous on $[a,b]$ and $N$ is between $f(a)$ and $f(b)$, then there exists $c \in (a,b)$ with $f(c) = N$.
• Squeeze Theorem: If $g(x) \leq f(x) \leq h(x)$ and $\lim g = \lim h = L$, then $\lim f = L$.

Key Limits

$\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$

Problem: Evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$.

Solution: Direct substitution gives $0/0$. Apply L'Hôpital's Rule: $\lim_{x \to 0} \frac{e^x}{1} = e^0 = 1$

• Try direct substitution first; use factoring, rationalization, or L'Hôpital's Rule for indeterminate forms.
• Continuity requires the function value to equal the limit.
• IVT and Squeeze Theorem are important for AP proofs/justifications.