MAT158H1/Unit 03
Unit 03 · Analysis 1A

Limits

The ε–δ definition, now as a tool you prove with, not just apply. In analysis every limit law is a theorem with a proof — this is where you learn to build those proofs yourself, including the standard tricks every analyst carries around.

By the end you can
Why it matters & where it comes from
Why it exists

For two centuries calculus rested on infinitesimals nobody could define. The ε–δ limit answered "what does approaches mean?" and made the whole subject rigorous.

In the real world

Every instantaneous rate is a limit, and the convergence of every numerical algorithm is a limit. The ε–δ habit of "given a target tolerance, find the needed precision" is exactly how error analysis works.

Story & open questions

Berkeley mocked infinitesimals in 1734 as "the ghosts of departed quantities." Cauchy and Weierstrass made ε–δ rigorous a century later; Weierstrass then built a function continuous everywhere yet differentiable nowhere.

The ε–δ definition

Definition Limit

$\displaystyle\lim_{x\to a}f(x)=L$ means: for every $\varepsilon>0$ there exists $\delta>0$ such that, for all $x$, $$0<|x-a|<\delta\ \Longrightarrow\ |f(x)-L|<\varepsilon.$$

Read it as a challenge–response game: the challenger picks a tolerance $\varepsilon$ around $L$, and you must produce a radius $\delta$ around $a$ that forces every $x$ (except possibly $a$ itself) to land within $\varepsilon$ of $L$. The "$0<|x-a|$" clause makes the value $f(a)$ — even its existence — irrelevant. Negation, for proving a limit is not $L$: there is an $\varepsilon>0$ such that every $\delta>0$ admits some $x$ with $0<|x-a|<\delta$ but $|f(x)-L|\ge\varepsilon$.

Example Quadratic

Prove $\lim_{x\to2}x^2=4$.

Scratch: $|x^2-4|=|x-2||x+2|$; restricting $\delta\le1$ keeps $x\in(1,3)$, so $|x+2|<5$.
Proof: let $\varepsilon>0$ and take $\delta=\min\{1,\varepsilon/5\}$. If $0<|x-2|<\delta$ then $|x^2-4|=|x-2||x+2|<5\cdot\tfrac\varepsilon5=\varepsilon$. ∎
Example A reciprocal

Prove $\lim_{x\to3}\dfrac1x=\dfrac13$.

Scratch: $\left|\tfrac1x-\tfrac13\right| =\dfrac{|x-3|}{3x}$. Restrict $\delta\le\tfrac32$ so $x>\tfrac32$, making $\tfrac{1}{3x}<\tfrac29$.
Proof: take $\delta=\min\{\tfrac32,\tfrac{9\varepsilon}2\}$. Then $\left|\tfrac1x-\tfrac13\right|<\tfrac29|x-3|<\tfrac29\cdot \tfrac{9\varepsilon}2=\varepsilon$. ∎
Scratch work versus proof Work backward to find $\delta$, then write the proof forward: announce $\delta$, verify it works. Restricting $\delta\le1$ first, to tame the factor you cannot control, is the single most reused trick in this course. Any valid $\delta$ is fine — it does not need to be the biggest one.

Uniqueness of limits

Theorem

A function has at most one limit at $a$: if $\lim_{x\to a}f=L$ and $\lim_{x\to a}f=M$, then $L=M$.

Proof
Suppose $L\ne M$ and set $\varepsilon=\tfrac{|L-M|}2>0$. Pick $\delta$ small enough that both $|f(x)-L|<\varepsilon$ and $|f(x)-M|<\varepsilon$ for $0<|x-a|<\delta$.
For any such $x$ (one exists — the domain contains points arbitrarily near $a$), the triangle inequality gives $|L-M|\le|L-f(x)|+|f(x)-M|<2\varepsilon=|L-M|$ — a contradiction. ∎

The limit laws, proved

Theorem

If $\lim_{x\to a}f=L$ and $\lim_{x\to a}g=M$, then $\lim(f+g)=L+M$, $\lim(fg)=LM$, and if $M\ne0$, $\lim\dfrac fg=\dfrac LM$.

Proof Sum law
Given $\varepsilon>0$, pick $\delta_1,\delta_2$ making $|f-L|<\tfrac\varepsilon2$ and $|g-M|<\tfrac\varepsilon2$ respectively.
For $\delta=\min\{\delta_1,\delta_2\}$: $|(f+g)-(L+M)|\le|f-L|+|g-M|<\varepsilon$. ∎
Proof idea Product law
Add and subtract: $fg-LM=f(g-M)+M(f-L)$.
Near $a$, $|f|\le|L|+1$ (take $\varepsilon=1$ in the definition). Then make $|g-M|<\tfrac{\varepsilon}{2(|L|+1)}$ and $|f-L|<\tfrac{\varepsilon}{2(|M|+1)}$; each product term is below $\tfrac\varepsilon2$. ∎
Proof idea Reciprocal
First force $|g(x)-M|<\tfrac{|M|}2$, which pins $|g(x)|>\tfrac{|M|}2$ — the denominator cannot collapse to $0$.
Then $\left|\tfrac1g-\tfrac1M\right| =\dfrac{|g-M|}{|g||M|}<\dfrac{2}{M^2}|g-M|$, and the right side is squeezed below $\varepsilon$ by the limit of $g$. The quotient law is product × reciprocal. ∎

The squeeze theorem, and limits that do not exist

Theorem Squeeze

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

Example

$\lim_{x\to0}x^2\sin(1/x)=0$: since $|\sin|\le1$, $-x^2\le x^2\sin(1/x)\le x^2$, and both bounds tend to $0$.

Example No limit at all

$\lim_{x\to0}\sin(1/x)$ does not exist.

Every interval $(0,\delta)$ contains points $x=\tfrac{1}{2\pi n+\pi/2}$ where $\sin(1/x)=1$ and points $x=\tfrac{1}{2\pi n-\pi/2}$ where $\sin(1/x)=-1$.
If the limit were $L$, taking $\varepsilon=1$ forces both $|1-L|<1$ and $|-1-L|<1$, i.e. $L>0$ and $L<0$ — impossible. ∎

One-sided limits and limits at infinity

$\lim_{x\to a^+}f=L$ uses $a<x<a+\delta$ in place of $0<|x-a|<\delta$ (similarly from the left); the two-sided limit exists iff both one-sided limits exist and agree. $\lim_{x\to\infty}f=L$ means: for every $\varepsilon>0$ there is $N$ with $x>N\Rightarrow|f(x)-L|<\varepsilon$. All the laws above carry over word for word.

Common pitfalls

• $\delta$ may depend on $\varepsilon$ and $a$ — never on $x$.
• Bound the uncontrolled factor (like $|x+2|$) by restricting $\delta\le1$ before solving for $\delta$.
• The limit ignores $f(a)$ entirely — that is the role of $0<|x-a|$.
• To disprove a limit, you get to pick $\varepsilon$; the opponent picks $\delta$; you must beat every $\delta$.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · Linear

Prove $\lim_{x\to1}(3x+2)=5$ from the definition.

Solution
Let $\varepsilon>0$ and take $\delta=\varepsilon/3$. If $0<|x-1|<\delta$ then $|(3x+2)-5|=3|x-1|<\varepsilon$. ∎
Practice 2 · Quadratic

Prove $\lim_{x\to2}(x^2+x)=6$ from the definition.

Solution
$|x^2+x-6|=|x-2||x+3|$. Restrict $\delta\le1$ so $|x+3|<6$; take $\delta=\min\{1,\varepsilon/6\}$. Then $|x^2+x-6|<6\cdot\tfrac\varepsilon6=\varepsilon$. ∎
Practice 3 · Squeeze

Find $\lim_{x\to0}x\cos(1/x)$ with justification.

Answer
$0$: since $-|x|\le x\cos(1/x)\le|x|$ and both bounds tend to $0$, the squeeze theorem applies.
Practice 4 · Square root

Prove $\lim_{x\to9}\sqrt{x}=3$ from the definition.

Solution
For $x\ge0$, $|\sqrt x-3|=\dfrac{|x-9|}{\sqrt x+3}\le\dfrac{|x-9|}3$. Take $\delta=\min\{9,3\varepsilon\}$ (the $9$ keeps $x>0$); then $|\sqrt x-3|<\tfrac{3\varepsilon}{3}=\varepsilon$. ∎
Practice 5 · Uniqueness

Write out the full proof that a limit, if it exists, is unique.

Solution
If $L\ne M$, set $\varepsilon=\tfrac{|L-M|}{2}$ and pick $\delta$ working for both limits. Any $x$ with $0<|x-a|<\delta$ gives $|L-M|\le|L-f(x)|+|f(x)-M|<|L-M|$, a contradiction.
Practice 6 · Product law

Prove the product law $\lim(fg)=LM$ in full.

Hint
$fg-LM=f(g-M)+M(f-L)$. Bound $|f|\le|L|+1$ near $a$, then make $|g-M|<\tfrac{\varepsilon} {2(|L|+1)}$ and $|f-L|<\tfrac{\varepsilon}{2(|M|+1)}$.
Practice 7 · Reciprocal law

Prove: if $\lim_{x\to a}g=M\ne0$, then $\lim_{x\to a}\dfrac1{g}= \dfrac1M$.

Solution
First choose $\delta_1$ so $|g-M|<\tfrac{|M|}2$, forcing $|g|>\tfrac{|M|}2$. Then $\left|\tfrac1g-\tfrac1M\right|=\tfrac{|g-M|}{|g||M|}< \tfrac{2}{M^2}|g-M|$; choose $\delta_2$ making $|g-M|<\tfrac{M^2\varepsilon}{2}$ and take the smaller δ.
Practice 8 · Non-existence

Prove that $\lim_{x\to0}\dfrac{|x|}{x}$ does not exist.

Solution
The function is $1$ for $x>0$ and $-1$ for $x<0$. Any candidate $L$ fails with $\varepsilon=1$: every $\delta$-interval contains points where the value is $1$ and points where it is $-1$, and no $L$ is within $1$ of both.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The limit spine is: quantify closeness in the range, choose closeness in the domain, then verify implication under punctured-neighborhood hypotheses.

Study routine. Before computing a limit, decide whether the task asks for proof, disproof, or use of a theorem; each has a different standard template.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Direct limit

Prove $\lim_{x o1}(2x+3)=5$.

Guide
Use $|(2x+3)-5|=2|x-1|$ and choose $\delta=\epsilon/2$.
Extension 2 · Product law idea

State the main bounding idea behind proving the product law for limits.

Guide
Write $fg-LM=f(g-M)+M(f-L)$ and bound $f$ near the point while making $g-M$ and $f-L$ small.
Extension 3 · Sequential disproof

Use sequences to show $\lim_{x o0}\sin(1/x)$ does not exist.

Guide
Choose $x_n=1/(\pi/2+2\pi n)$ giving value $1$, and $y_n=1/(3\pi/2+2\pi n)$ giving value $-1$. Both approach $0$.
Extension 4 · One-sided

Why does $\lim_{x o0} |x|/x$ not exist?

Guide
The right-hand limit is $1$ and the left-hand limit is $-1$, so the two-sided limit fails.
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