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The definition, the order of its quantifiers, the restrict-then-bound technique, and how a limit is disproved.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the epsilon-delta definition with its quantifiers in the right order, explain what the $0 < |x - a|$ excludes and why the derivative needs it excluded, find a $\delta$ for a linear limit directly and for a quadratic one by restricting first and bounding second, write the proof forwards with the scratch work discarded, and disprove a limit by naming a single $\varepsilon$ that defeats every $\delta$ and every candidate value at once.
The last lesson said a limit means the values get "as close as you like" to $L$ by taking $x$ "close enough" to $a$. Both phrases are doing real work and neither has been defined. As close as you like is a challenge someone else issues; close enough is an answer you supply after hearing it. The definition below is those two sentences with the vagueness removed and the order of play written down.
$$\lim_{x \to a} f(x) = L \iff \forall \varepsilon > 0\ \exists \delta > 0 : 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$$
Read it as a game. A challenger names a tolerance $\varepsilon$ — how close to $L$ the outputs must be. You answer with a radius $\delta$ — how close to $a$ the inputs must be. You win the round if every $x$ within $\delta$ of $a$, except $a$ itself, has $f(x)$ within $\varepsilon$ of $L$. The limit is $L$ if you can win every round.
Three details carry the whole definition.
The order. $\varepsilon$ first, then $\delta$. $\delta$ is allowed to depend on $\varepsilon$, and always does except for constant functions. Swapping the quantifiers gives a statement that is not weaker but different, and almost never true.
The $0 <$. The hypothesis excludes $x = a$, so the function's value there — or its absence — is not part of the claim. This is what makes $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ a sensible thing to write.
The implication runs one way. You are promised nothing about $x$ outside the window, and nothing about whether $f$ ever equals $L$.
To disprove a limit, negate it: there is an $\varepsilon > 0$ such that for every $\delta > 0$ some $x$ with $0 < |x - a| < \delta$ has $|f(x) - L| \ge \varepsilon$. One bad $\varepsilon$ is a complete proof.
Another way: picture
A horizontal band of half-width $\varepsilon$ around $L$ and a vertical band of half-width $\delta$ around $a$. The definition says the graph over the vertical band — with the single line $x = a$ removed — stays inside the horizontal band. A challenger narrowing the horizontal band forces you to narrow the vertical one; a limit exists when you can always narrow it enough.
Another way: steps
The proof is found backwards and written forwards.
Found: 1. Write $|f(x) - L|$ and factor out $|x - a|$. 2. Bound the other factor by restricting to $|x - a| < 1$. 3. Set $\delta = \min\bigl(1, \varepsilon / \text{bound}\bigr)$.
Written: "Let $\varepsilon > 0$. Put $\delta = \ldots$. Then $0 < |x - a| < \delta$ gives $\ldots < \varepsilon$." The scratch work never appears; the reader only has to check that the stated $\delta$ works.
For a linear function the factor multiplying $|x - a|$ is a constant and the proof is one line. For anything else it is a function of $x$, and you cannot divide $\varepsilon$ by something that varies.
The fix is to shrink the playing field first. Restricting to $|x - a| < 1$ pins $x$ to an interval, and on an interval the varying factor has a constant bound. Then $\delta = \min(1, \varepsilon/\text{bound})$ honours both constraints at once, and taking the smaller of two working values is always safe.
For $\lim_{x \to 1} x^2 = 1$: $|x^2 - 1| = |x-1|\,|x+1|$. On $|x - 1| < 1$ we have $0 < x < 2$, so $|x + 1| < 3$, so $|x^2 - 1| < 3|x-1|$. Take $\delta = \min(1, \varepsilon/3)$.
The $1$ is not sacred — $\tfrac12$ or $10$ would do, with a different bound. What matters is that some restriction is imposed before the varying factor is bounded.
Producing a $\delta$ that depends on $x$. $\delta$ may depend on $\varepsilon$ and on $a$. It may not depend on $x$: $x$ is quantified inside the implication, and by the time $x$ is being discussed $\delta$ has already been chosen. A "proof" ending in $\delta = \varepsilon / |x + 1|$ is not a proof.
Bounding the varying factor without restricting first. "$|x + 1| < 3$" is not true for all $x$. It is true on the interval the restriction created, and the restriction has to be stated.
Proving the converse. Showing that $|f(x) - L| < \varepsilon$ implies $|x - a| < \delta$ is a different statement and proves nothing.
Disproving by naming the wrong limit. To show no limit exists you must defeat every candidate $L$, not just the obvious one — which is why the two-approaches argument is the standard tool: it produces two values that no single $L$ can be close to at once.
Thinking $\delta$ must be largest. Any $\delta$ that works is a complete answer. Only a question that asks for the largest one is asking for it.
Nothing in $\forall \varepsilon\ \exists \delta$ tells you what $L$ is. The definition verifies a limit you already suspect; it never produces one. That is why the last lesson's algebra comes first — factor, cancel, rationalise to see what the value must be — and this lesson's machinery comes second, to prove that the value is right.
Which is also why almost nobody computes limits this way in practice. The next lesson proves the limit laws once, using this definition, and after that the laws do the work and the definition is what the laws stand on. The definition is load-bearing and rarely handled, like a foundation.
Prove $\lim_{x \to 2} (3x - 1) = 5$. Scratch: $|3x - 1 - 5| = |3x - 6| = 3|x - 2|$.
Factor out $|x - a|$.
To make $3|x-2| < \varepsilon$ it is enough that $|x - 2| < \varepsilon/3$.
Solve for the input bound.
Proof. Let $\varepsilon > 0$ and put $\delta = \varepsilon/3$. If $0 < |x - 2| < \delta$ then $|(3x - 1) - 5| = 3|x - 2| < 3\delta = \varepsilon$. $\blacksquare$
Written forwards, with the scratch work thrown away.
Claim: $\lim_{x \to 0} \sin(1/x)$ does not exist. Suppose it were $L$ and take $\varepsilon = \tfrac12$.
Fix one challenge.
Any $\delta > 0$ admits $x$ with $1/x = \pi/2 + 2k\pi$ (where $\sin = 1$) and $x$ with $1/x = -\pi/2 + 2k\pi$ (where $\sin = -1$), by taking $k$ large.
Every window, however small, contains both.
$L$ would have to be within $\tfrac12$ of both $1$ and $-1$, and no number is. So no $L$ works and the limit does not exist. Notice the shape: the argument defeats every candidate at once.
$|x^2 - 2x - 3| = |x - 3|\,|x + 1|$, so restrict: on $|x - 3| < 1$, $2 < x < 4$ and $|x + 1| < 5$.
Restrict, then bound.
So $|x^2 - 2x - 3| < 5|x - 3|$; choose $\delta = \min(1, \varepsilon/5)$.
Written forwards: let $\varepsilon > 0$, put $\delta = \min(1, \varepsilon/5)$, and the two constraints give the result.
Match each proof step to its role.
| Begins with the requested output precision | Provides a response tolerance | Introduces any allowed nearby input | Establishes the definition's conclusion | |
|---|---|---|---|---|
| Choose an arbitrary $\varepsilon>0$ | ||||
| Set $\delta=\varepsilon/3$ | ||||
| Assume $0<|x-a|<\delta$ | ||||
| Show $|f(x)-L|<3\delta=\varepsilon$ |
Match each epsilon-delta phrase to its logical role.
| The challenger requests output accuracy | The proof supplies an input tolerance | Any permitted input close to $a$ | The resulting output is close to $L$ | |
|---|---|---|---|---|
| For every $\varepsilon>0$ | ||||
| There exists $\delta>0$ | ||||
| $0<|x-a|<\delta$ | ||||
| $|f(x)-L|<\varepsilon$ |
Match each part of an epsilon–delta proof to the role it plays.
| An arbitrary requested output tolerance | An input tolerance that may depend on epsilon | The hypothesis that an input is close but not equal | The guaranteed output closeness | |
|---|---|---|---|---|
| For every epsilon greater than zero | ||||
| There exists delta greater than zero | ||||
| $0<|x-a|<\delta$ | ||||
| $|f(x)-L|<\varepsilon$ |
For $f(x) = 5x + 1$ and $\varepsilon = 2/10$, what is the largest $\delta$ that makes $0 < |x - 2| < \delta$ force $|f(x) - f(2)| < \varepsilon$?
Answer:
In the definition of $\lim_{x \to a} f(x) = L$, which order do the quantifiers come in?
$\lim_{x \to \infty} \dfrac{2}{x} = 0$ means: for every $\varepsilon > 0$ there is an $M$ with $x > M$ forcing $\left|\dfrac{2}{x}\right| < \varepsilon$. For $\varepsilon = 1/10$, what is the smallest such $M$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = 4x - 4$ and $|x - 3| < 1/5$. What bound on $|f(x) - f(3)|$ does that guarantee?
Answer:
You can find and write an epsilon-delta proof, and disprove a limit with one $\varepsilon$. Say in your own words why $\delta$ may depend on $\varepsilon$ but never on $x$. Next: building a $\delta$ when the factor to bound is not a constant.
9. Your turn: prove $\lim_{x \to 3} (x^2 - 2x) = 3$, step 3