The conceptual heart of the course. The ε–δ definition of a limit is the single idea everything downstream — derivatives, integrals, series — is built on. Spend real time until you can write ε–δ proofs cold.
Early calculus worked remarkably well, but many infinitesimal arguments lacked a precise foundation. The ε–δ limit answered the question — what does "approaches" actually mean? — and gave analysis a rigorous framework.
Every instantaneous rate is a limit: speed on a speedometer, a tolerance an engineer designs to, whether an algorithm settles on an answer. Continuity is the hidden assumption behind root-finding by bisection and behind "small change in input, small change in output" in physics and economics.
Newton and Leibniz built calculus in the 1600s on infinitesimals; Bishop Berkeley ridiculed them in 1734 as "the ghosts of departed quantities." It took Cauchy and Weierstrass another century to make ε–δ rigorous. Weierstrass then exhibited a function that is continuous everywhere but differentiable nowhere — destroying everyone's intuition at once.
An $\varepsilon$-$\delta$ proof is a game with two distances. The output distance $|f(x)-L|$ is what the challenger wants small. The input distance $|x-a|$ is what you control. Scratch work moves backward from the output inequality to a condition on $|x-a|$; the final proof then moves forward from your chosen $\delta$ to the desired output bound.
Not every limit needs a full $\varepsilon$-$\delta$ proof. The course expects you to know when a theorem is doing the work and when the definition is being tested directly.
$\displaystyle\lim_{x\to a}f(x)=L$ 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 $L$; you must produce a radius $\delta$ around $a$ so that every $x$ within $\delta$ (except $a$ itself) lands within $\varepsilon$ of $L$. The "$0<|x-a|$" part says the value at $a$ is irrelevant. The order of quantifiers matters: $\delta$ may depend on $\varepsilon$, but it must work for every eligible $x$.
Move ε. The red horizontal band is the required output tolerance; the blue vertical band is the input window produced by the proof's conservative choice $\delta=\min\{1,\varepsilon/5\}$.
The structure is always the same: "Let $\varepsilon>0$. Choose $\delta=\dots$. Then if $0<|x-a|<\delta$, …, so $|f(x)-L|<\varepsilon$."
Prove $\lim_{x\to3}(2x-1)=5$.
Prove $\lim_{x\to2}x^2=4$.
If $\lim_{x\to a}f=L$ and $\lim_{x\to a}g=M$, then the limits of $f\pm g$, $fg$, and (if $M\ne0$) $f/g$ are $L\pm M$, $LM$, and $L/M$.
If $g(x)\le f(x)\le h(x)$ for all $x$ sufficiently close to $a$ (possibly excluding $a$), and $\lim_{x\to a}g=\lim_{x\to a}h=L$, then $\lim_{x\to a}f=L$.
$\lim_{x\to0}x^2\sin(1/x)=0$, since $-x^2\le x^2\sin(1/x)\le x^2$ and both bounds tend to $0$.
$\lim_{x\to a^-}$ and $\lim_{x\to a^+}$ restrict $x$ to one side; the two-sided limit exists iff both one-sided limits exist and agree. Infinite limits and limits at infinity use the same challenge–response structure.
$\lim_{x\to a}f(x)=\infty$ means: for every $M>0$, there is a $\delta>0$ such that $0<|x-a|<\delta\Rightarrow f(x)>M$.
$\lim_{x\to\infty}f(x)=L$ means: for every $\varepsilon>0$, there is an $N>0$ such that $x>N\Rightarrow |f(x)-L|<\varepsilon$.
$f$ is continuous at $a$ if $a$ is in the domain and $\displaystyle\lim_{x\to a}f(x)=f(a)$. Equivalently: $\forall\varepsilon>0\, \exists\delta>0$ with $|x-a|<\delta\Rightarrow|f(x)-f(a)|<\varepsilon$.
Sums, products, quotients (nonzero denominator), and compositions of continuous functions are continuous. Polynomials are continuous everywhere; rational functions wherever defined.
Let $f(x)=x\sin(1/x)$ for $x\ne0$ and $f(0)=0$. Is $f$ continuous at $0$?
If $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then there is some $c\in[a,b]$ with $f(c)=N$.
$p(x)=x^3+x-1$ is continuous, $p(0)=-1$, and $p(1)=1$. By the IVT, some $c\in(0,1)$ satisfies $p(c)=0$. The theorem proves existence, not a formula for $c$.
Discontinuities commonly appear as removable holes, jump discontinuities where one-sided limits disagree, or infinite discontinuities. The ε–δ definition handles all of them without relying on a graph.
• Choosing $\delta$ that secretly depends on $x$ — it may depend only on
$\varepsilon$ (and $a$).
• Forgetting to bound the "extra" factor (like $|x+2|$) by restricting $\delta\le
1$ first.
• Confusing "limit exists" with "function is defined": the limit ignores the
value at $a$.
Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.
Let $\varepsilon>0$. Find all $\delta>0$ such that $|x-3|<\delta\Rightarrow|5x-15|<\varepsilon$.
Prove $\lim_{x\to3}(5x+1)=16$ directly from the definition.
Prove $\lim_{x\to0}(x^3+x^2)=0$ directly from the definition.
Prove: if $\lim_{x\to a}|f(x)|=0$, then $\lim_{x\to a}f(x)=0$.
Suppose $\lim_{x\to a}f(x)=0$ and $g$ is bounded near $a$. Prove $\lim_{x\to a}f(x)g(x)=0$.
Compute $\lim_{x\to\infty}\dfrac{x^2+2x+3}{3x^2+4x+5}$.
Using $\lim_{x\to0}\dfrac{\sin x}{x}=1$, compute $\lim_{x\to0}\dfrac{\sin(5x)}{x}$.
Prove that $x^4-2x=100$ has at least two solutions.
Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:
Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.
The core chain is: limit definition gives pointwise control; continuity transfers nearby input control to output control; compact-interval theorems give existence.
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.
Prove from the definition that $\lim_{x o2}(3x-1)=5$.
Show that $f(x)=x^2$ is continuous at $a$ using epsilon-delta reasoning.
Prove that $x^3+x-1=0$ has a root in $(0,1)$.
Give a function with $\lim_{x o0} f(x)$ not existing even though both one-sided limits are finite.