Chapter 01

Rigorous

Limits and Continuity

00 · Symbol Glossary

$\mathbb{R}$Blackboard R

The set of all real numbers — every number on the number line: negative, zero, positive, decimals, irrationals like $\pi$ . The double-struck style distinguishes it from an ordinary letter $R$ . Every function and limit in this chapter operates on real numbers.

$f(x)$f of x — function

A rule that assigns exactly one output to each input $x$ . Read "f of x." The letter $f$ names the function; $x$ is the input variable. $f(3)$ means "evaluate $f$ at $x=3$ ."

$\lim_{x \to a} f(x)$limit of f of x as x approaches a

The value that $f(x)$ gets arbitrarily close to as $x$ gets arbitrarily close to $a$ — without necessarily ever equaling $a$ . This is the central object of the chapter. Read aloud: "the limit as x approaches a of f of x."

$\to$approaches / tends to

An arrow meaning "gets arbitrarily close to, without necessarily reaching." $x \to a$ means $x$ moves toward $a$ from either or both sides, never required to land exactly on $a$ .

$x \to a^+$x approaches a from the right

The right-hand limit: $x$ approaches $a$ only through values greater than $a$ . The superscript $+$ marks "from above."

$x \to a^-$x approaches a from the left

The left-hand limit: $x$ approaches $a$ only through values less than $a$ . The superscript $-$ marks "from below."

$\varepsilon$epsilon — target tolerance

Greek lowercase epsilon. In the formal definition of a limit, $\varepsilon$ is an arbitrarily small positive number that measures how close $f(x)$ must be to the limit value. Chosen first, by whoever is testing the limit.

$\delta$delta — input tolerance

Greek lowercase delta. A positive number measuring how close $x$ must be to $a$ to guarantee $f(x)$ lands within $\varepsilon$ of the limit. Chosen second, in response to $\varepsilon$ , and typically depends on it.

$\forall,\ \exists$"for all", "there exists"

Quantifiers borrowed from formal logic. $\forall\,\varepsilon>0$ reads "for every positive $\varepsilon$ , no matter how small." $\exists\,\delta>0$ reads "there exists some positive $\delta$ ." Together they build the precise limit definition in Section 02.

$\lvert x-a \rvert$absolute value of x minus a — distance

The distance between $x$ and $a$ on the number line, always non-negative. $\lvert x-a\rvert < \delta$ means " $x$ is within $\delta$ of $a$ ."

$\infty,\ -\infty$infinity, negative infinity

Not real numbers — symbols describing unbounded growth. $\lim_{x\to a}f(x)=\infty$ means $f(x)$ grows without bound as $x\to a$ ; it does not mean the limit exists as a number. $x\to\infty$ means $x$ grows without bound.

$\text{DNE}$does not exist

Shorthand written when a limit fails to settle on a single finite value — the left- and right-hand limits disagree, or the function oscillates, or it grows without bound.

01 · The Idea of a Limit

Suppose $f(x) = \dfrac{x^2-1}{x-1}$ . Plug in $x=1$ directly and you get $\dfrac{0}{0}$ — undefined. But look at what happens as $x$ gets close to $1$ without touching it: at $x=0.9$ , $f(0.9)=1.9$ ; at $x=0.99$ , $f(0.99)=1.99$ ; at $x=1.01$ , $f(1.01)=2.01$ . The outputs are closing in on $2$ from both sides, even though $f(1)$ itself is undefined.

That closing-in value is the limit. A limit describes the trend of a function near a point, independent of whether the function is even defined at that point.

Definition — Limit (informal)

We write

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

if $f(x)$ gets arbitrarily close to the number $L$ whenever $x$ gets sufficiently close to $a$ , from either side, with $x \neq a$ .

$L$ — the limit value, a single real number (when the limit exists).

$x \neq a$ — the behavior at $a$ is irrelevant; only the behavior near $a$ matters.

Example — Evaluating a limit by direct substitution

For a well-behaved function like $f(x) = 3x + 2$ , the limit as $x \to 4$ is found by simply plugging in:

\lim_{x \to 4} (3x+2) = 3(4)+2 = 14

The $12$ came from $3\times4$ ; the final $14$ from $12+2$ . This works because $f$ is a polynomial — polynomials have no gaps, jumps, or holes anywhere, so the limit always equals the direct evaluation.

❌ Direct substitution gives 0/0

For $f(x) = \dfrac{x^2-1}{x-1}$ and $a=1$ : substituting gives $\dfrac{1^2-1}{1-1} = \dfrac{0}{0}$ .

Why it breaks: $\frac{0}{0}$ is an indeterminate form — it does not mean the limit is $0$ , or $1$ , or undefined outright. It means direct substitution gives no information and a different method is required.

Correct approach: factor first. $x^2-1=(x-1)(x+1)$ , so for $x\neq1$ : $\dfrac{(x-1)(x+1)}{x-1} = x+1$ . Since the limit only cares about $x$ near $1$ , not at $1$ , the cancellation is valid, and $\lim_{x\to1}(x+1) = 2$ — matching the numerical trend from above.

Consequence: whenever substitution yields $\frac{0}{0}$ , algebraic simplification (factoring, rationalizing) must be tried before concluding anything.

02 · The Formal $\varepsilon$ – $\delta$ Definition

The informal phrase "gets arbitrarily close" needs a precise meaning to be usable in proofs. The formal definition pins it down with two tolerances: how close the output must be ( $\varepsilon$ ), and how close the input must be to guarantee it ( $\delta$ ).

Definition — Limit (formal, epsilon-delta)

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

\forall\, \varepsilon > 0,\ \exists\, \delta > 0 \text{ such that } 0 < \lvert x-a\rvert < \delta \implies \lvert f(x)-L\rvert < \varepsilon

Read left to right: "for every tolerance $\varepsilon>0$ someone hands you, you can find a tolerance $\delta>0$ such that whenever $x$ is within $\delta$ of $a$ (but not equal to $a$ ), $f(x)$ is guaranteed to land within $\varepsilon$ of $L$ ."

$0 < \lvert x-a \rvert$ — strictly positive, meaning $x \neq a$ ; the point $a$ itself is excluded.

Step-by-step — Prove $\lim_{x\to3}(2x-1)=5$ using epsilon-delta

Start from the target inequality: we need

\lvert f(x)-L\rvert < \varepsilon

, i.e.

\lvert (2x-1)-5\rvert < \varepsilon

Simplify the expression inside the bars:

(2x-1)-5 = 2x-6 = 2(x-3)

, so the condition becomes

\lvert 2(x-3)\rvert < \varepsilon

, i.e.

2\lvert x-3\rvert < \varepsilon

Isolate $\lvert x-3\rvert$ : divide both sides by

2

\lvert x-3\rvert < \varepsilon/2

Choose $\delta$ : set

\delta = \varepsilon/2

. This choice guarantees that whenever

0<\lvert x-3\rvert<\delta

, reversing steps 3–2 gives

\lvert f(x)-5\rvert<\varepsilon

Numeric check: if

\varepsilon = 0.1

, then

\delta = 0.05

. Take

x = 3.03

(within

\delta

3

, since

\lvert3.03-3\rvert=0.03<0.05

). Then

f(3.03) = 2(3.03)-1 = 5.06

, and

\lvert5.06-5\rvert=0.06<0.1

. ✓

Why bother with epsilon-delta if the answer is obvious?

For a straight line, the limit is visually obvious. The formal definition matters because it is the only tool precise enough to handle functions where intuition fails — oscillating functions, piecewise functions, and rigorous proofs of the limit laws used constantly from here on.

03 · One-Sided Limits and Existence

A limit is a single statement about behavior from both directions at once. Sometimes the trend from the left disagrees with the trend from the right — in that case, the two-sided limit does not exist, even though each one-sided limit exists individually.

Definition — One-Sided and Two-Sided Limits

$\lim_{x\to a^-}f(x)=L_1$ — the value $f(x)$ approaches as $x$ approaches $a$ using only values less than $a$ .

$\lim_{x\to a^+}f(x)=L_2$ — the value $f(x)$ approaches using only values greater than $a$ .

\lim_{x\to a} f(x) \text{ exists} \iff \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x)

The two-sided limit exists if and only if both one-sided limits exist and agree.

Example — Step function with a jump

Let $f(x) = \begin{cases} 1 & x < 0 \\ 3 & x \geq 0 \end{cases}$ .

From the left: $\lim_{x\to0^-}f(x) = 1$ (values just below $0$ all give output $1$ ).

From the right: $\lim_{x\to0^+}f(x) = 3$ (values just above $0$ all give output $3$ ).

Since $1 \neq 3$ , $\lim_{x\to0}f(x)$ does not exist. This is a jump discontinuity — revisited in Section 05.

❌ Assuming a limit exists because f(a) is defined

For $f(x) = \begin{cases} x^2 & x \neq 2 \\ 10 & x = 2 \end{cases}$ , a student might reason: " $f(2)=10$ is defined, so the limit at $2$ is $10$ ."

Why it breaks: the limit only cares about values near $2$ , not the value assigned at $2$ . As $x\to2$ from either side, $f(x)=x^2 \to 4$ , since the definition $f(x)=10$ only kicks in exactly at $x=2$ , a single point that the approaching values never touch.

Correct value: $\lim_{x\to2}f(x) = 4 \neq f(2) = 10$ .

Consequence: the limit and the function value at a point are two separate quantities. They coincide for continuous functions (Section 05) but can disagree elsewhere, as here.

04 · Limit Laws and Computing Limits

Once a few basic limits are known, complicated limits can be built up algebraically rather than re-proving epsilon-delta every time.

Definition — Limit Laws

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

\lim_{x\to a}\big[f(x)\pm g(x)\big] = L \pm M \qquad \lim_{x\to a}\big[c\cdot f(x)\big] = cL

\lim_{x\to a}\big[f(x)g(x)\big] = LM \qquad \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M} \ \ (M\neq0)

$c$ — any constant (real number), not depending on $x$ .

Step-by-step — Evaluate $\lim_{x\to2}\dfrac{x^3-4x}{x+1}$

Check the denominator's limit first:

\lim_{x\to2}(x+1) = 2+1=3 \neq 0

, so the quotient law applies directly — no factoring needed.

Evaluate the numerator's limit:

\lim_{x\to2}(x^3-4x) = 2^3 - 4(2) = 8-8=0

. The

8

comes from

2^3=2\times2\times2

; the second

8

from

4\times2

Apply the quotient law:

\dfrac{0}{3}=0

Conclude:

\lim_{x\to2}\dfrac{x^3-4x}{x+1} = 0

. Note this is different from the earlier

\frac00

case — here the numerator is

0

but the denominator is not, which is a perfectly valid limit of

0

, not an indeterminate form.

05 · Continuity

A function is continuous at a point if its graph has no hole, jump, or break there — you could trace it through that point without lifting a pen. This intuition is formalized by tying together the limit and the function value.

Definition — Continuity at a Point

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

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

If $f$ is continuous at every point of an interval, it is continuous on that interval.

Example — Verifying continuity

$f(x) = x^2 + 1$ at $a = 3$ : $f(3) = 10$ (defined). $\lim_{x\to3}(x^2+1) = 3^2+1=10$ (exists). Both equal $10$ . All three conditions hold — continuous at $x=3$ .

Three flavors of discontinuity

Removable: the limit exists but disagrees with (or is missing from) $f(a)$ — a single-point hole, as in Section 03's $x^2$ example. Jump: the one-sided limits exist but disagree, as in the step function above. Infinite: $f(x)\to\pm\infty$ near $a$ , typically from a zero denominator — covered next.

❌ Infinite discontinuity at a vertical asymptote

For $f(x) = \dfrac{1}{x-2}$ at $a=2$ : $f(2)$ is undefined (division by $0$ ). Checking the trend: as $x\to2^+$ , $x-2$ is a tiny positive number, so $f(x)\to+\infty$ . As $x\to2^-$ , $x-2$ is a tiny negative number, so $f(x)\to-\infty$ .

Why it breaks: the one-sided trends do not approach a finite number — they grow without bound in opposite directions. Neither $\lim_{x\to2^-}f(x)$ nor $\lim_{x\to2^+}f(x)$ is a real number, so condition 2 of continuity fails outright, and the two-sided limit does not exist.

Consequence: $f$ has an infinite discontinuity at $x=2$ , marked graphically by a vertical asymptote — the curve never crosses the vertical line $x=2$ and shoots off toward $\pm\infty$ on either side of it.

06 · The Intermediate Value Theorem

Continuity has a powerful consequence: a continuous function cannot skip over values. If it starts below a target height and ends above it, it must cross that height somewhere in between.

Definition — Intermediate Value Theorem (IVT)

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

$[a,b]$ — the closed interval from $a$ to $b$ , including both endpoints.

Example — Locating a root with the IVT

Let $f(x) = x^3 - x - 1$ . This is a polynomial, so it is continuous everywhere. Compute $f(1) = 1-1-1=-1$ and $f(2) = 8-2-1=5$ .

Since $f$ is continuous on $[1,2]$ and $0$ lies between $-1$ and $5$ , the IVT guarantees some $c \in [1,2]$ with $f(c)=0$ — a root of $f$ exists in that interval, even without solving for it exactly. This is the theoretical justification behind numerical root-finding methods used throughout applied mathematics.

❌ Applying the IVT to a discontinuous function

Let $f(x) = \dfrac{1}{x-1}$ on $[0,2]$ , with $f(0)=-1$ and $f(2)=1$ . A student might claim: " $0$ is between $-1$ and $1$ , so by the IVT some $c\in[0,2]$ gives $f(c)=0$ ."

Why it breaks: $f$ is not continuous on $[0,2]$ — it has an infinite discontinuity at $x=1$ (Section 05), so the hypothesis of the IVT is not satisfied.

Correct conclusion: in fact $\frac{1}{x-1}=0$ has no solution at all — the numerator $1$ can never equal $0$ . The IVT's guarantee simply does not apply here.

Consequence: always verify continuity on the entire interval before invoking the IVT — checking only the endpoints is not sufficient.

07 · Exercises

EXERCISE 1.1

Try direct substitution first. If it gives $\frac{0}{0}$ , factor the numerator and cancel the common factor with the denominator.

Direct substitution gives $\dfrac{3^2-9}{3-3}=\dfrac{0}{0}$ , indeterminate. Factor: $x^2-9=(x-3)(x+3)$ , so for $x\neq3$ : $\dfrac{(x-3)(x+3)}{x-3}=x+3$ . Then $\lim_{x\to3}(x+3)=3+3=6$ .

Evaluate $\lim_{x\to3}\dfrac{x^2-9}{x-3}$ . Show the indeterminate form and the factoring step explicitly.

EXERCISE 1.2

Compute the left-hand and right-hand limits separately by plugging into each piece of the definition, then compare.

$f(x)=\begin{cases}2x+1 & x<1\\5 & x=1\\x^2+2 & x>1\end{cases}$ . Left: $\lim_{x\to1^-}(2x+1)=2(1)+1=3$ . Right: $\lim_{x\to1^+}(x^2+2)=1^2+2=3$ . Both equal $3$ , so $\lim_{x\to1}f(x)=3$ . But $f(1)=5\neq3$ , so the limit exists yet $f$ is not continuous at $x=1$ (a removable discontinuity).

For $f(x)=\begin{cases}2x+1 & x<1\\5 & x=1\\x^2+2 & x>1\end{cases}$ , find $\lim_{x\to1}f(x)$ and determine whether $f$ is continuous at $x=1$ .

EXERCISE 1.3

Use the formal definition: simplify $\lvert f(x)-L\rvert < \varepsilon$ down to an inequality in $\lvert x-a\rvert$ , then read off $\delta$ in terms of $\varepsilon$ .

Need $\lvert(4x-3)-5\rvert<\varepsilon \iff \lvert4x-8\rvert<\varepsilon \iff 4\lvert x-2\rvert<\varepsilon \iff \lvert x-2\rvert<\varepsilon/4$ . Choose $\delta=\varepsilon/4$ . Check with $\varepsilon=0.08$ : $\delta=0.02$ . At $x=2.01$ : $f(2.01)=4(2.01)-3=5.04$ , and $\lvert5.04-5\rvert=0.04<0.08$ . ✓

Using the epsilon-delta definition, find $\delta$ in terms of $\varepsilon$ to prove $\lim_{x\to2}(4x-3)=5$ . Verify numerically with $\varepsilon=0.08$ .

EXERCISE 1.4

Identify what kind of expression sits in the denominator near the point in question, and check the one-sided trends separately.

$f(x)=\dfrac{x+3}{x^2-9}=\dfrac{x+3}{(x-3)(x+3)}$ . At $x=-3$ : for $x\neq-3$ , this simplifies to $\dfrac{1}{x-3}$ , which is defined and finite at $x=-3$ (value $-\frac14$ ) — so $x=-3$ is a removable discontinuity (a hole), not infinite. At $x=3$ : the simplified form $\frac{1}{x-3}$ blows up as $x\to3$ , giving $+\infty$ from the right and $-\infty$ from the left — an infinite discontinuity.

For $f(x)=\dfrac{x+3}{x^2-9}$ , classify the discontinuities at $x=-3$ and at $x=3$ as removable, jump, or infinite. Justify each with the appropriate limit computation.

EXERCISE 1.5

Confirm $f$ is continuous on the whole interval, evaluate both endpoints, and check whether $0$ lies between them.

$f(x)=x^3-3x+1$ is a polynomial, hence continuous everywhere, including on $[0,2]$ . $f(0)=0-0+1=1$ . $f(2)=8-6+1=3$ . Since $0$ is not between $1$ and $3$ , the IVT as stated does not directly guarantee a root on $[0,2]$ using these endpoints — check $f(1)=1-3+1=-1$ instead: $f(0)=1>0$ and $f(1)=-1<0$ , and $0$ lies between $1$ and $-1$ , so by the IVT there is a root in $(0,1)$ .

Show that $f(x)=x^3-3x+1$ has a root somewhere in the interval $(0,1)$ using the Intermediate Value Theorem.

EXERCISE 1.6

A quoted price feed that "jumps" between two levels with no intermediate value is the discrete analogue of a function with a jump discontinuity.

Model the fee schedule as $f(x)=\begin{cases}0.10x & 0\le x<10{,}000\\0.08x & x\ge10{,}000\end{cases}$ where $x$ is trade size. Left-hand limit at $x=10{,}000$ : $0.10(10{,}000)=1{,}000$ . Right-hand limit: $0.08(10{,}000)=800$ . Since $1{,}000\neq800$ , the two one-sided limits disagree — the fee function has a jump discontinuity at $x=10{,}000$ : a trade of $9{,}999$ costs more in fees than a trade of $10{,}000$ , a discontinuous (and exploitable) jump in the schedule.

A trading fee schedule charges $10\%$ on trade sizes under $10{,}000$ dollars and $8\%$ on trade sizes at or above $10{,}000$ dollars. Model this as a piecewise function of trade size $x$ , and show it has a jump discontinuity at $x=10{,}000$ by computing both one-sided limits there.

08 · Chapter Summary

Concept	Formula / Rule
Limit (informal)	$f(x)\to L$ as $x\to a$ , $x\neq a$
Limit (formal)	$\forall\varepsilon>0\,\exists\delta>0: 0<\lvert x-a\rvert<\delta\implies\lvert f(x)-L\rvert<\varepsilon$
One-sided limits	$\lim_{x\to a}f(x)$ exists $\iff$ left limit = right limit
Limit laws	Sum, product, quotient limits combine from known limits
Indeterminate form	$\frac{0}{0}$ : substitution fails; factor or simplify first
Continuity at $a$	$f(a)$ defined, limit exists, and they're equal
Removable discontinuity	Limit exists but $\neq f(a)$ (or $f(a)$ undefined)
Jump discontinuity	One-sided limits exist but disagree
Infinite discontinuity	One-sided limit(s) are $\pm\infty$ ; vertical asymptote
IVT	Continuous on $[a,b]$ $\implies$ hits every value between $f(a)$ and $f(b)$

Next: Chapter 02 — Derivatives: Definition and Rules, where the limit machinery built here is used to define the instantaneous rate of change.