Limits and Continuity

What Is a Limit?

The limit of $f(x)$ as $x$ approaches $a$ is the value that $f(x)$ gets arbitrarily close to as $x$ gets arbitrarily close to $a$ (from both sides), regardless of what happens at $x = a$ itself.

We write:

$\lim_{x \to a} f(x) = L$

This means: for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$. You won't need to write epsilon-delta proofs on the AP exam, but understanding this definition helps you reason about limits correctly.

Limit Laws

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

• Sum/Difference Law: $\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$
• Constant Multiple Law: $\lim_{x \to a} [c \cdot f(x)] = c \cdot L$
• Product Law: $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
• Quotient Law: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided $M \neq 0$
• Power Law: $\lim_{x \to a} [f(x)]^n = L^n$ for any positive integer $n$
• Root Law: $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$, provided $L > 0$ when $n$ is even

These laws allow you to evaluate most limits by direct substitution when the function is continuous at $a$.

Direct Substitution

The simplest way to evaluate a limit: plug in $x = a$. This works whenever $f$ is continuous at $a$. For example:

$\lim_{x \to 3} (2x^2 - 5x + 1) = 2(9) - 15 + 1 = 4$

Direct substitution works for all polynomials, rational functions (where the denominator is nonzero), trigonometric functions at points in their domain, exponential and logarithmic functions at points in their domain, and any composition or combination of the above (by the limit laws).

Indeterminate Forms and Algebraic Techniques

When direct substitution gives $\frac{0}{0}$, the limit may still exist — you just need to do more work. Common techniques include:

Factoring: Factor numerator and denominator, then cancel the common factor.

$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x + 2) = 4$

Rationalizing: Multiply by the conjugate when you see square roots.

$\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} = \lim_{x \to 0} \frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{4}$

Simplifying complex fractions: Combine fractions in the numerator or denominator into a single fraction, then simplify.

$\lim_{x \to 0} \frac{\frac{1}{x+3} - \frac{1}{3}}{x} = \lim_{x \to 0} \frac{\frac{3 - (x+3)}{3(x+3)}}{x} = \lim_{x \to 0} \frac{-x}{3x(x+3)} = \lim_{x \to 0} \frac{-1}{3(x+3)} = -\frac{1}{9}$

Special Trigonometric Limits

Two limits you must memorize:

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

These are used constantly. For example:

$\lim_{x \to 0} \frac{\sin(5x)}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin(5x)}{5x} = 5 \cdot 1 = 5$

One-Sided Limits

The left-hand limit $\lim_{x \to a^-} f(x)$ considers only values of $x$ less than $a$. The right-hand limit $\lim_{x \to a^+} f(x)$ considers only values greater than $a$.

The two-sided limit $\lim_{x \to a} f(x) = L$ exists if and only if both one-sided limits exist and are equal:

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$

One-sided limits are especially important for piecewise functions. For example, if:

$f(x) = \begin{cases} x^2 & x < 1 \\ 2x - 1 & x \geq 1 \end{cases}$

Then $\lim_{x \to 1^-} f(x) = 1$ and $\lim_{x \to 1^+} f(x) = 1$, so $\lim_{x \to 1} f(x) = 1$.

Continuity

A function $f$ is continuous at $x = a$ if all three conditions hold:

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

If any condition fails, $f$ has a discontinuity at $a$. Types of discontinuities:

• Removable (hole): The limit exists but either $f(a)$ is undefined or $f(a) \neq \lim_{x \to a} f(x)$. Example: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.
• Jump: Both one-sided limits exist but are not equal. Example: the greatest integer function $\lfloor x \rfloor$ at any integer.
• Infinite (vertical asymptote): At least one one-sided limit is $\pm \infty$. Example: $f(x) = \frac{1}{x}$ at $x = 0$.

The Intermediate Value Theorem (IVT)

If $f$ is continuous on $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a, b)$ such that $f(c) = N$.

The IVT is commonly used to show that an equation has a solution. If $f(1) = -2$ and $f(3) = 5$ and $f$ is continuous, then there must be some $c \in (1, 3)$ where $f(c) = 0$.

The Squeeze Theorem

If $g(x) \leq f(x) \leq h(x)$ near $a$ (except possibly at $a$) and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$.

Classic example: $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$, since $-x^2 \leq x^2 \sin(1/x) \leq x^2$ and both bounds approach 0.

Tip 1: Always Try Direct Substitution First

Plug in the value. If you get a real number, that's your answer. Only if you get $\frac{0}{0}$ (or another indeterminate form) do you need algebraic manipulation.

Tip 2: Recognize the Form Before Choosing a Technique

If you get $\frac{0}{0}$ from a rational expression, try factoring. If the expression involves a square root, try rationalizing (multiply by the conjugate). If you see $\frac{\sin(\text{something})}{\text{something}}$, aim to use the special trig limit.

Tip 3: For Piecewise Functions, Check Both Sides

Always compute left-hand and right-hand limits separately at the boundary. If they match, the two-sided limit exists. Then check all three continuity conditions.

Tip 4: Don't Confuse "Limit Exists" with "Function Is Continuous"

A limit can exist at a point even if the function is undefined there (that's a removable discontinuity). Continuity requires the limit to exist AND equal the function value.

Tip 5: Use the Squeeze Theorem When Oscillation Is Involved

Whenever you see $\sin(1/x)$ or $\cos(1/x)$ multiplied by something going to zero, the Squeeze Theorem is usually the right approach.