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For every $\varepsilon$ a $\delta$: the deleted neighbourhood, the two-stage estimate, and the sequential criterion.
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 prove a limit of a function from the epsilon-delta definition, running the two-stage estimate when a loose factor has to be bounded first and naming the delta as a minimum. You will also be able to state the sequential criterion, use it to carry sequence theorems to functions, and use it backwards — with two sequences — to prove that a limit does not exist.
The epsilon-N definition for sequences, and the two-stage estimate from the absolute value lesson. The definition below is the sequence definition with the index replaced by a distance, and the estimates are the same estimates.
The deleted neighbourhood of $c$ of radius $\delta$ is the set of $x$ with $0 < |x - c| < \delta$: near $c$, and not $c$. A limit is only asked for at a limit point of the domain — a point every deleted neighbourhood of which meets the domain — because otherwise the condition is vacuous and every $L$ would qualify.
> $\lim_{x \to c} f(x) = L$ means: for every $\varepsilon > 0$ there is $\delta > 0$ such that $0 < |x - c| < \delta$ implies $|f(x) - L| < \varepsilon$.
The challenge and response are as before. The challenge $\varepsilon$ is an accuracy in the output; the response $\delta$ is a tolerance in the input that guarantees it.
The point $c$ is excluded. That is what $0 < |x - c|$ says, and it is not a technicality: it is what lets $\frac{x^2 - 4}{x - 2}$ have the limit $4$ at $x = 2$ where it is not even defined. A limit describes the neighbourhood, never the point.
The sequential criterion. $\lim_{x \to c} f(x) = L$ exactly when $f(x_n) \to L$ for every sequence $x_n \to c$ with $x_n \ne c$.
This is the bridge back to unit 2. It transports the algebra of limits, the squeeze theorem and every other sequence result to functions at once, and — read backwards — it is the standard way to prove a limit does not exist: exhibit two sequences whose images go to different places.
One-sided limits. $\lim_{x \to c^+}$ restricts to $x > c$, $\lim_{x \to c^-}$ to $x < c$. The two-sided limit exists exactly when both exist and agree.
A convergence proof is not an observation that the terms get close. It is a rule that turns a given $\varepsilon$ into an $N$, and the rule is the proof: no $N$, no theorem.
Another way: picture
Draw a horizontal band of half-width $\varepsilon$ about the height $L$. The claim is that some vertical strip about $c$, with the line above $c$ itself removed, has its whole graph inside the band. Narrow the band and the strip must usually narrow too; the limit exists when a strip can always be found.
Another way: steps
To prove a limit from the definition:
Step 2 is the draft and step 4 is the proof. Handing in the draft is the commonest way to lose the argument, because the implication then runs the wrong way.
For $f(x) = x^2$ at $c$: $|x^2 - c^2| = |x - c|\,|x + c|$, and $|x+c|$ is not controlled by $|x - c|$.
Stage one. Insist $\delta \le 1$. Then $|x - c| < 1$ gives $|x| < |c| + 1$, so $|x + c| < 2|c| + 1$.
Stage two. Now $|x^2 - c^2| < (2|c|+1)|x-c|$, so $|x - c| < \dfrac{\varepsilon}{2|c|+1}$ finishes.
Name it. $\delta = \min\left(1, \dfrac{\varepsilon}{2|c|+1}\right)$.
The minimum is not timidity. Stage one is what made the constant in stage two available, so both restrictions must hold at once, and a single $\delta$ that enforces both is exactly their minimum. The same structure appears for every non-linear function: a crude restriction to bound the loose factor, then a fine one in terms of $\varepsilon$.
| At $c$ | Limit exists | Value defined | They agree |
|---|---|---|---|
| $f(x) = x^2$ at $2$ | yes, $4$ | yes, $4$ | yes — continuous |
| $\dfrac{x^2-4}{x-2}$ at $2$ | yes, $4$ | no | not applicable — removable |
| $f(x) = \lfloor x \rfloor$ at $1$ | no (sides differ) | yes, $1$ | no — jump |
| $\sin(1/x)$ at $0$ | no | no | no — oscillation |
Every row is a different situation, and the definition is deliberately blind to the second column. Continuity, next lesson, is precisely the assertion that the first three columns all say yes.
Using $f(c)$ as evidence. The definition excludes $x = c$. A limit can exist where the value is missing, and can differ from the value where it is present.
Choosing $\delta$ before $\varepsilon$. The order of the quantifiers is the content. A $\delta$ that did not depend on $\varepsilon$ would say the function is constant near $c$.
Handing in the scratch work. The exploration runs from $\varepsilon$ backwards to $\delta$; the proof runs from the assumption $|x - c| < \delta$ forwards to the conclusion. The two are not the same argument written differently.
Forgetting the minimum. A delta from stage two alone does not enforce the restriction stage one needed, so the constant used in stage two was never justified.
Claim: $\lim_{x \to 1}(5x + 3) = 8$. Let $\varepsilon > 0$.
The challenge is fixed first.
Scrap: $|5x + 3 - 8| = 5|x - 1|$, which is below $\varepsilon$ when $|x - 1| < \varepsilon/5$.
Work backwards, on scrap.
Proof: take $\delta = \varepsilon/5$. If $0 < |x - 1| < \delta$ then $|5x + 3 - 8| = 5|x-1| < 5\delta = \varepsilon$.
Write it forwards, with the delta named.
Claim: $\lim_{x \to 0} \dfrac{|x|}{x}$ does not exist.
A claim of non-existence.
Take $x_n = 1/n \to 0$: the images are all $1$. Take $x_n = -1/n \to 0$: the images are all $-1$.
Two sequences, both approaching the point.
The sequential criterion asks that every such sequence give the same limit, and these give $1$ and $-1$. So no $L$ works.
One criterion, applied backwards.
Scrap: $\left|\dfrac{1}{x} - \dfrac{1}{3}\right| = \dfrac{|3 - x|}{3|x|}$, and $|x|$ in the denominator is the loose factor.
Find the factor that is not controlled.
Stage one: $\delta \le 1$ gives $x \ge 2$, so the quantity is at most $|x - 3|/6$.
A crude restriction, keeping $x$ away from zero.
So $\delta = \min(1, 6\varepsilon)$ works, and the proof is written forwards from there. The first restriction is essential rather than cosmetic: without it $x$ could approach $0$, where the quantity is unbounded however small the other restriction is made.
Let $f(x) = 6x + 3$, so $\lim_{x \to 1} f(x) = 9$. Give the set of positive $\delta$ for which $0 < |x - 1| < \delta$ guarantees $|f(x) - 9| < 4$.
This task has no paper form; do it on a device.
Let $f(x) = x^2$ and $c = 3$, with $\varepsilon = 8$. Using the standard two-stage estimate, what delta does the recipe $\delta = \min\left(1, \dfrac{\varepsilon}{2c + 1}\right)$ give?
Answer:
Put the steps of the proof that $\lim_{x \to 1} (9x + 3) = 12$ into the order they are written in.
Number the steps in order (write the number in the box):
Match each statement about $f$ at the point $8$ to what it asserts.
| the values near the point approach $L$, with the point itself excluded | the limit exists, the value exists, and they agree | the values approach $L$ from above the point only | the value at the point, with no claim about nearby values | |
|---|---|---|---|---|
| $\lim_{x \to 8} f(x) = L$ | ||||
| $f$ is continuous at $8$ | ||||
| $\lim_{x \to 8^{+}} f(x) = L$ | ||||
| $f(8) = L$ |
Let $f(x) = \dfrac{x^2 - 4^2}{x - 4}$ for $x \ne 4$, and $f(4) = 0$. Here is an argument about $\lim_{x \to 4} f(x)$. Mark the one step that is not sound.
This task has no paper form; do it on a device.
Show that $f(x) = \sin(1/x)$ has no limit as $x \to 0$. Which piece of evidence does that?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $f(x) = 9x + 3$, so $\lim_{x \to 1} f(x) = 12$. Give the set of positive $\delta$ for which $0 < |x - 1| < \delta$ guarantees $|f(x) - 12| < 8$.
This task has no paper form; do it on a device.
You can produce a delta for a given epsilon, write an epsilon-delta proof forwards rather than as scratch work, and deny a limit with two sequences. Say in your own words why the definition excludes the point itself. Next: what is added when the limit is required to equal the value.
10. Your turn: prove $\lim_{x \to 3} \dfrac{1}{x} = \dfrac{1}{3}$, step 3