Mathematics / Calculus I Single-Variable Differential Calculus 100% Free Open Access
Chapter 3 • Theory & Derivations

The Theory of Limits & Rigorous ε-δ Analysis

The mathematical foundation of single-variable analysis: intuitive limit mechanics, left and right one-sided limits, Cauchy-Weierstrass epsilon-delta definition and scratchwork methodology, limit arithmetic laws, Squeeze theorem, geometric proof of lim (sin theta)/theta = 1, limits at infinity, and infinite limits with asymptotes.

§3.1 The Intuitive Limit Concept, Numerical Behavior & One-Sided Limits

1. Intuitive Limit Concept

In calculus, the limit is the foundational concept that enables the transition from static algebra to dynamical rates of change. Intuitively, we say that the limit of $f(x)$ as $x$ approaches $c$ is $L$:

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

if the values of $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $c$ (with $x \ne c$). Crucially, the value of $f(c)$ itself is completely irrelevant to the limit; $f(c)$ may equal $L$, may equal some other value, or may be completely undefined.

2. One-Sided Limits

Often, a function behaves differently depending on whether $x$ approaches $c$ from numbers strictly less than $c$ (from the left) or strictly greater than $c$ (from the right):

$$\begin{aligned} \mathbf{\text{Left-Hand Limit: }} & \lim_{x \to c^-} f(x) = L_1 \quad (\text{as } x \to c \text{ with } x < c) \ \mathbf{\text{Right-Hand Limit: }} & \lim_{x \to c^+} f(x) = L_2 \quad (\text{as } x \to c \text{ with } x > c) \end{aligned}$$

Theorem (Two-Sided Limit Existence Criterion): The two-sided limit $\lim_{x \to c} f(x)$ exists and equals $L$ if and only if both one-sided limits exist and are strictly equal:

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

If $\lim_{x \to c^-} f(x) \ne \lim_{x \to c^+} f(x)$, the two-sided limit does not exist (DNE).

§3.2 The Formal Cauchy-Weierstrass Epsilon-Delta Definition of Limit

1. The Formal Cauchy-Weierstrass Definition

To eliminate ambiguity in phrases like "arbitrarily close" and "sufficiently close", Augustin-Louis Cauchy and Karl Weierstrass formulated the rigorous $\varepsilon$-$\delta$ definition:

$$\mathbf{\lim_{x \to c} f(x) = L \iff \forall \varepsilon > 0, \, \exists \delta > 0 \text{ such that } 0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon}$$

2. Geometric Interpretation of the $\varepsilon$-$\delta$ Challenge

The definition is a mathematical game between two players:

  1. A challenger proposes a tolerance $\varepsilon > 0$, demanding that $f(x)$ lie in the horizontal target band $(L - \varepsilon, L + \varepsilon)$.
  2. You must respond by finding a neighborhood radius $\delta > 0$ such that for every $x$ within the punctured vertical interval $(c - \delta, c + \delta) \setminus \{c\}$, the graph $y = f(x)$ stays strictly inside the horizontal $\varepsilon$-tube.

Notice that $\delta$ depends on both $\varepsilon$ and $c$: $\delta = \delta(\varepsilon, c)$.

3. The Canonical $\varepsilon$-$\delta$ Proof Template

A rigorous $\varepsilon$-$\delta$ proof always proceeds in two distinct phases:

  • Scratchwork Analysis: Start with the target inequality $|f(x) - L| < \varepsilon$. Factor out $|x - c|$ to obtain $|x - c| \cdot |g(x)| < \varepsilon$. Choose a preliminary bound on $|x - c|$ (typically $\delta_1 \le 1$) to bound the auxiliary factor $|g(x)| \le M$. Then deduce $\delta = \min\{1, \varepsilon / M\}$.
  • Formal Deductive Proof: State "Let $\varepsilon > 0$ be given. Choose $\delta = \min\{1, \varepsilon / M\}$." Then, assuming $0 < |x - c| < \delta$, deduce sequentially that $|f(x) - L| < \varepsilon$ by direct algebraic inequalities.

§3.3 Limit Arithmetic Laws & The Squeeze (Sandwich) Theorem

1. Limit Arithmetic Theorems

Let $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$. Then:

$$\begin{aligned} \text{Sum / Difference: } & \lim_{x \to c} [f(x) \pm g(x)] = L \pm M \ \text{Scalar Multiple: } & \lim_{x \to c} [k f(x)] = k L \quad (k \in \mathbb{R}) \ \text{Product Law: } & \lim_{x \to c} [f(x) g(x)] = L \cdot M \ \text{Quotient Law: } & \lim_{x \to c} \left[\frac{f(x)}{g(x)}\right] = \frac{L}{M} \quad (\text{provided } M \ne 0) \ \text{Power / Root Law: } & \lim_{x \to c} [f(x)]^n = L^n, \quad \lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L} \quad (L > 0 \text{ if } n \text{ is even}) \end{aligned}$$

2. The Squeeze (Sandwich) Theorem

$$\mathbf{\text{Theorem: If } g(x) \le f(x) \le h(x) \text{ for all } x \text{ in an open interval containing } c \text{ (except possibly at } c\text{)}, \text{ and } \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, \text{ then } \lim_{x \to c} f(x) = L.}$$

3. Rigorous Geometric Proof of the Fundamental Limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$

Consider a unit circle of radius $r = 1$. For an acute central angle $\theta \in (0, \pi/2)$ measured in radians:

  1. Area of inner triangle with vertices $(0,0), (1,0), (\cos\theta, \sin\theta)$:
    $$A_{\text{inner}} = \frac{1}{2} (1) (\sin\theta) = \frac{1}{2} \sin\theta$$
  2. Area of circular sector of angle $\theta$:
    $$A_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \theta$$
  3. Area of outer tangent triangle with vertices $(0,0), (1,0), (1, \tan\theta)$:
    $$A_{\text{outer}} = \frac{1}{2} (1) (\tan\theta) = \frac{1}{2} \frac{\sin\theta}{\cos\theta}$$

By geometric containment, $A_{\text{inner}} < A_{\text{sector}} < A_{\text{outer}}$:

$$\frac{1}{2} \sin\theta < \frac{1}{2} \theta < \frac{1}{2} \frac{\sin\theta}{\cos\theta}$$

Multiplying by $2 / \sin\theta$ (since $\sin\theta > 0$ on $(0, \pi/2)$):

$$1 < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta} \implies \cos\theta < \frac{\sin\theta}{\theta} < 1$$

Taking the limit as $\theta \to 0^+$: since $\lim_{\theta \to 0^+} \cos\theta = 1$, the Squeeze Theorem forces $\lim_{\theta \to 0^+} \frac{\sin\theta}{\theta} = 1$. By even symmetry $\frac{\sin(-\theta)}{-\theta} = \frac{\sin\theta}{\theta}$, the two-sided limit is established: $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \quad \blacksquare$

§3.4 Limits at Infinity, Infinite Limits & Asymptotic Analysis

1. Limits at Infinity & Horizontal Asymptotes

$$\mathbf{\lim_{x \to \infty} f(x) = L \iff \forall \varepsilon > 0, \, \exists N > 0 \text{ such that } x > N \implies |f(x) - L| < \varepsilon}$$

If $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$, the horizontal line $y = L$ is a horizontal asymptote of the graph $y = f(x)$.

2. Fundamental Rational Limit Theorem

For any rational power $r > 0$:

$$\lim_{x \to \infty} \frac{1}{x^r} = 0, \qquad \lim_{x \to -\infty} \frac{1}{x^r} = 0 \quad (\text{for } x^r \text{ defined for negative } x)$$

3. Infinite Limits & Vertical Asymptotes

$$\mathbf{\lim_{x \to c} f(x) = \infty \iff \forall M > 0, \, \exists \delta > 0 \text{ such that } 0 < |x - c| < \delta \implies f(x) > M}$$

If $f(x) \to \pm\infty$ as $x \to c^+$ or $x \to c^-$, the line $x = c$ is a vertical asymptote.

TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 3.1: Direct Algebraic Limit Evaluation with Indeterminate 0/0 Conjugation

Evaluate the exact analytical limit: $\lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2}$. Show all factorization and rationalization steps.

Step 1: Check Direct Substitution
$$\frac{\sqrt{0^2 + 9} - 3}{0^2} = \frac{3 - 3}{0} = \left[\frac{0}{0}\right]$$

Direct substitution yields the indeterminate form $0/0$, necessitating algebraic transformation.

Step 2: Multiply by Radical Conjugate
$$\frac{\sqrt{x^2 + 9} - 3}{x^2} \cdot \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} = \frac{(x^2 + 9) - 9}{x^2 (\sqrt{x^2 + 9} + 3)} = \frac{x^2}{x^2 (\sqrt{x^2 + 9} + 3)}$$

Expanding the difference of squares in the numerator eliminates the radical.

Step 3: Cancel Common Zero Factor and Evaluate
$$\lim_{x \to 0} \frac{1}{\sqrt{x^2 + 9} + 3} = \frac{1}{\sqrt{0 + 9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$$

Since $x e 0$ in the punctured limit neighborhood, we cancel $x^2$ and evaluate by direct substitution.

Final Answer & Physical Insight

\lim_{x \to 0} \frac{\sqrt{x^2 + 9} - 3}{x^2} = \frac{1}{6}

Tier 2: Intermediate Exam Example 3.2: Rigorous Cauchy Epsilon-Delta Proof for a Quadratic Function

Use the formal $\varepsilon$-$\delta$ definition of a limit to prove rigorously that $\lim_{x \to 3} (2x^2 - 5x + 1) = 4$.

Step 1: Scratchwork Analysis
$$|(2x^2 - 5x + 1) - 4| = |2x^2 - 5x - 3| = |(2x + 1)(x - 3)| = |2x + 1| \cdot |x - 3| < \varepsilon$$

We factored out the critical deviation factor $|x - 3|$. We must now bound the auxiliary factor $|2x + 1|$.

Step 2: Establish a Preliminary Bound on |x - 3|
$$\text{Assume } |x - 3| < 1 \implies -1 < x - 3 < 1 \implies 2 < x < 4$$

Bounding $x$ on $(2, 4)$ allows us to bound $2x + 1$:

Step 3: Bound the Auxiliary Term
$$2 < x < 4 \implies 4 < 2x < 8 \implies 5 < 2x + 1 < 9 \implies |2x + 1| < 9$$

Thus, whenever $|x - 3| < 1$, we have $|2x + 1| \cdot |x - 3| < 9 |x - 3|$. To ensure this is $< \varepsilon$, we require $|x - 3| < \varepsilon / 9$.

Step 4: Formal Proof
$$\text{Let } \varepsilon > 0. \text{ Choose } \delta = \min\left(1, \frac{\varepsilon}{9}\right). \text{ If } 0 < |x - 3| < \delta, \text{ then } |(2x^2 - 5x + 1) - 4| < 9|x - 3| < 9\left(\frac{\varepsilon}{9}\right) = \varepsilon. \quad \blacksquare$$

This completes the formal deductive proof according to Cauchy's criterion.

Final Answer & Physical Insight

\delta = \min\left(1, \frac{\varepsilon}{9}\right) \implies |(2x^2 - 5x + 1) - 4| < \varepsilon

Tier 3: Honors / Proof Challenge Example 3.3: Trigonometric Squeeze Theorem Proof for an Oscillatory Radical Limit

Evaluate the limit $\lim_{x \to 0} \frac{x^3 \sin\left(\frac{1}{x}\right) + \tan(4x)}{\sin(2x) + x^2 \cos\left(\frac{1}{x^2}\right)}$ using rigorous limit theorems and the Squeeze Theorem. Prove every step without L'Hôpital's Rule.

Step 1: Divide Numerator and Denominator by $x$
$$\frac{x^3 \sin(1/x) + \tan(4x)}{\sin(2x) + x^2 \cos(1/x^2)} = \frac{x^2 \sin(1/x) + \frac{\tan(4x)}{x}}{\frac{\sin(2x)}{x} + x \cos(1/x^2)}$$

Dividing by $x e 0$ isolates the fundamental trigonometric limits and small oscillatory terms.

Step 2: Evaluate the Squeezed Oscillatory Limits
$$-1 \le \sin(1/x) \le 1 \implies -x^2 \le x^2 \sin(1/x) \le x^2 \implies \lim_{x \to 0} x^2 \sin(1/x) = 0$$

Similarly, $-|x| \le x \cos(1/x^2) \le |x| \implies \lim_{x o 0} x \cos(1/x^2) = 0$ by the Squeeze Theorem.

Step 3: Evaluate the Trigonometric Limits
$$\lim_{x \to 0} \frac{\tan(4x)}{x} = \lim_{x \to 0} 4 \left(\frac{\sin(4x)}{4x}\right) \frac{1}{\cos(4x)} = 4(1)(1) = 4, \qquad \lim_{x \to 0} \frac{\sin(2x)}{x} = 2(1) = 2$$

Using the fundamental theorem $\lim_{u o 0} \frac{\sin u}{u} = 1$.

Step 4: Combine via Quotient and Sum Limit Laws
$$\lim_{x \to 0} \frac{x^2 \sin(1/x) + \frac{\tan(4x)}{x}}{\frac{\sin(2x)}{x} + x \cos(1/x^2)} = \frac{0 + 4}{2 + 0} = \frac{4}{2} = 2$$

Since the denominator limit $2 e 0$, the quotient limit law holds rigorously.

Final Answer & Physical Insight

\lim_{x \to 0} \frac{x^3 \sin(1/x) + \tan(4x)}{\sin(2x) + x^2 \cos(1/x^2)} = 2