1.1 Limits, Continuity, and Epsilon-Delta Definitions

1.1 Limits, Continuity, and Epsilon-Delta Definitions

Differential calculus rests upon the concept of the limit, establishing the rigorous analytical foundation for continuity and differentiability. On the GRE Mathematics Subject Test, questions test formal $\varepsilon$-$\delta$ mechanics, topological implications of continuity, and the precise classification of discontinuities.

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The Formal Epsilon-Delta Definition of a Limit

Let $f: D \to \mathbb{R}$ with accumulation point $a$ of $D$. We say $\lim_{x \to a} f(x) = L$ if: $\forall \varepsilon > 0, \; \exists \delta > 0 \text{ such that } \forall x \in D, \; 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$

Key Structural Rules

• Quantifier Hierarchy: Tolerance $\varepsilon > 0$ is given first. Radius $\delta > 0$ depends on $\varepsilon$ and $a$, denoted $\delta(\varepsilon, a)$. In uniform continuity, $\delta$ depends solely on $\varepsilon$.
• Punctured Neighborhood: The condition $0 < |x - a|$ excludes $x = a$. The value $f(a)$ need not exist and does not affect $L$.
• One-Sided Limits: The right limit $\lim_{x \to a^+} f(x) = L$ restricts inputs to $x > a$; the left limit $\lim_{x \to a^-} f(x) = L$ restricts to $x < a$. The two-sided limit exists if and only if both exist and agree: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$.
• Infinite Limits: $\lim_{x \to a} f(x) = \infty \iff \forall M > 0, \exists \delta > 0$ such that $0 < |x - a| < \delta \implies f(x) > M$.

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Continuity, IVT, and EVT

A function $f$ is continuous at $a$ if $\lim_{x \to a} f(x) = f(a)$, requiring three criteria:

1. $f(a)$ is defined ($a \in \operatorname{Dom}(f)$).
2. $\lim_{x \to a} f(x)$ exists in $\mathbb{R}$.
3. $\lim_{x \to a} f(x) = f(a)$.

Intermediate Value Theorem (IVT)

If $f: [a, b] \to \mathbb{R}$ is continuous and $u$ lies between $f(a)$ and $f(b)$, there exists $c \in (a, b)$ with $f(c) = u$. Continuous maps preserve connectedness.

• Root Finding: If $f(a) f(b) < 0$ on $[a, b]$, then $f(c) = 0$ for some $c \in (a, b)$.

Extreme Value Theorem (EVT)

If $K \subset \mathbb{R}$ is compact (closed and bounded) and $f: K \to \mathbb{R}$ is continuous, then $f(K)$ is compact. Thus $f$ attains an absolute maximum $M$ and minimum $m$ on $K$: $\exists x_{\min}, x_{\max} \in K \quad \text{such that} \quad f(x_{\min}) \le f(x) \le f(x_{\max}) \quad \forall x \in K.$

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Classification of Discontinuities

Discontinuities at $x = a$ are classified by one-sided limits:

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The Squeeze Theorem and Standard Asymptotics

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

Standard First-Order Limits

1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
2. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
3. $\lim_{x \to 0} \frac{\tan x}{x} = 1, \quad \lim_{x \to 0} \frac{\arctan x}{x} = 1$
4. $\lim_{x \to 0} |x|^\alpha \sin(1/x) = 0$ for any $\alpha > 0$

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Step-by-Step Worked Problems

Problem 1: Rigorous $\varepsilon$-$\delta$ Verification

Prove $\lim_{x \to 2} (3x^2 - 5x) = 2$ using the $\varepsilon$-$\delta$ definition.

Solution:

1. Target error: $|(3x^2 - 5x) - 2| = |(3x + 1)(x - 2)| = |3x + 1| \cdot |x - 2|$.
2. Set $\delta_0 = 1$. Then $|x - 2| < 1 \implies 1 < x < 3$, so $4 < 3x + 1 < 10$, giving $|3x + 1| < 10$.
3. Thus $|f(x) - 2| < 10 |x - 2|$. To ensure this is $< \varepsilon$, require $|x - 2| < \varepsilon / 10$.
4. Choose $\delta = \min(1, \varepsilon / 10)$. If $0 < |x - 2| < \delta$, then $|(3x^2 - 5x) - 2| < 10(\varepsilon / 10) = \varepsilon$.

Problem 2: Squeeze Theorem Evaluation

Compute $\lim_{x \to 0} \frac{x^2 \sin(1/x) + 3 \sin(2x)}{2x + \tan(4x)}$.

Solution:

1. Divide numerator and denominator by $x$: $\frac{x \sin(1/x) + 6 \cdot \frac{\sin(2x)}{2x}}{2 + 4 \cdot \frac{\tan(4x)}{4x}}.$
2. Because $|\sin(1/x)| \le 1$, $-|x| \le x \sin(1/x) \le |x|$, yielding $\lim_{x \to 0} x \sin(1/x) = 0$ by the Squeeze Theorem.
3. The numerator approaches $0 + 6(1) = 6$ and the denominator approaches $2 + 4(1) = 6$.
4. The limit evaluates to $6 / 6 = 1$.

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GRE Exam Traps & Pitfalls

Trap 1: Applying EVT on Non-Closed or Unbounded Sets EVT fails without compactness. The function $f(x) = 1/x$ on $(0, 1]$ is continuous on a bounded set but unbounded near $0$. Similarly, $f(x) = x$ on $[0, \infty)$ is continuous on a closed set but has no maximum.

Trap 2: Concluding Continuity From Limit Existence Limit existence $\lim_{x \to a} f(x) = L$ does not guarantee continuity. If $f(a)$ is undefined or $f(a) \neq L$, $f$ has a removable discontinuity at $a$.

Trap 3: Reversed Epsilon-Delta Quantifiers Beware the statement "$\exists \delta > 0, \forall \varepsilon > 0$". That reverses quantifiers and asserts a single $\delta$ works for all $\varepsilon$, which forces $f$ to be locally constant.