Back to the on-screen lesson ·
Restrict, bound the varying factor, divide and take the minimum — the proof for a limit that is not linear.
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 non-linear limit from the definition: factor $|x - a|$ out of $|f(x) - L|$, choose a restriction that keeps the remaining factor bounded — including the case where it is a denominator that must be kept away from zero — find a bound on the interval the restriction creates, and assemble $\delta = \min(c, \varepsilon/K)$ knowing what each argument of the minimum is protecting. You will also be able to split a target between two error terms, which is the shape of every convergence proof that follows.
For a linear limit the proof is one line: $|f(x) - L|$ is a constant times $|x - a|$, and the constant divides into $\varepsilon$. For anything else the factor multiplying $|x - a|$ varies with $x$, and you cannot divide $\varepsilon$ by a moving target.
The standard construction has three steps and they always come in this order.
The $\min$ is the step that is most often dropped and least often understood. Without it a large $\varepsilon$ produces a $\delta$ bigger than $c$, the restriction is abandoned, and $K$ — proved only on the restricted interval — no longer holds. The proof collapses precisely where nobody looks.
Choosing $c$. Take $c = 1$ unless the function misbehaves within $1$ of $a$. For $1/x$ near $a > 0$ the denominator can reach $0$, so take $c = a/2$: that forces $x > a/2$ and keeps the denominator away from zero. The rule is: choose $c$ small enough that the varying factor is bounded on the interval it creates.
Another way: picture
Two nested windows round $a$. The outer one, of radius $c$, is where the bound $K$ was proved; the inner one, of radius $\varepsilon/K$, is where the conclusion needs $x$ to be. $\delta$ is whichever is smaller, so that $x$ is always inside both — and if the inner window were allowed to stick out past the outer one, the argument inside it would be quoting a bound that was never proved there.
Another way: steps
Step 3 is mechanical once the interval is written down.
For $|x + a|$ near $a$ with $c = 1$: $x \in (a - 1, a + 1)$, so $x + a \in (2a - 1, 2a + 1)$, so $|x + a| < 2a + 1$ — take the end of largest magnitude, and take it strictly since the interval is open.
For $|x^2 + ax + a^2|$ near $a$ with $c = 1$, each term is bounded separately and the triangle inequality adds them: $|x^2| < (|a|+1)^2$, $|ax| < |a|(|a|+1)$, so $K = (|a|+1)^2 + |a|(|a|+1) + a^2$. Not the tightest bound available, and it does not need to be: any valid $K$ gives a valid proof. Chasing the smallest one is wasted effort.
For a factor in a denominator, the bound runs the other way: you need the denominator to stay large, so bound it below, and the reciprocal is then bounded above.
Dropping the $\min$. The most common error, and invisible for small $\varepsilon$, which is what makes it dangerous: the proof looks right on every example anyone tries.
Bounding on the wrong interval. The bound must be proved on the interval the restriction creates, not asserted generally. $|x + 1| < 3$ is false for $x = 10$ and true on $|x - 1| < 1$, and which of those you mean has to be written down.
Letting $\delta$ mention $x$. $\delta$ is chosen before $x$ is considered. A formula for $\delta$ containing $x$ is not an answer to the challenge.
Restricting with a $c$ that does not tame the function. For $1/x$ near $a = 0.5$, taking $c = 1$ admits $x$ arbitrarily close to $0$ and the factor is unbounded on the interval. The restriction has to be chosen for the function, not by habit.
A great deal of effort goes into finding the largest $\delta$, or the tightest bound $K$, and none of it is required. The definition asks whether a $\delta$ exists; producing one — however wasteful — settles the question. A bound of $19$ where $13$ would have done makes the proof no weaker and costs nothing.
The one place tightness matters is a question that asks for it, and those questions are asking something different: not "does the limit hold" but "how close must the input be", which is a quantitative question about the function rather than a proof about the limit.
Prove $\lim_{x \to 2} x^3 = 8$. Factor: $|x^3 - 8| = |x - 2|\,|x^2 + 2x + 4|$.
The difference of cubes does the factoring.
Restrict to $|x - 2| < 1$, so $1 < x < 3$. Then $|x^2| < 9$, $|2x| < 6$ and the constant is $4$, so $|x^2 + 2x + 4| < 19$.
Bound each term on the interval and add.
Take $\delta = \min(1, \varepsilon/19)$. Then $0 < |x - 2| < \delta$ gives $|x^3 - 8| < 19\,|x - 2| < \varepsilon$. $\blacksquare$
$19$ is generous; generous is fine.
Prove $\lim_{x \to 3} \dfrac{1}{x} = \dfrac13$. Combine: $\left|\dfrac1x - \dfrac13\right| = \dfrac{|3 - x|}{3|x|} = \dfrac{|x-3|}{3|x|}$.
The varying factor is in the denominator.
Restrict to $|x - 3| < \tfrac32$, so $x > \tfrac32$. Then $3|x| > \tfrac92$, so $\dfrac{1}{3|x|} < \dfrac{2}{9}$.
Bound the denominator below to bound the fraction above.
Take $\delta = \min\bigl(\tfrac32,\ \tfrac92 \varepsilon\bigr)$. Why $\tfrac32$ and not $1$? Because the restriction has to keep $x$ away from $0$, and for a point nearer the origin than $3$ the radius $1$ would not.
The restriction is a choice, made for a reason.
$|x^2 - 16| = |x - 4|\,|x + 4|$. Restrict to $|x - 4| < 1$: then $3 < x < 5$.
Restrict.
So $|x + 4| < 9$, and $|x^2 - 16| < 9|x - 4|$.
Bound.
$\delta = \min(1, \varepsilon/9)$.
Match each estimate to the proof decision it supports.
| Exposes the distance from the limit point | Keeps the auxiliary factor controllable | Gives a fixed multiplier for the error | Makes both proof requirements hold | |
|---|---|---|---|---|
| $|x^2-9|=|x-3||x+3|$ | ||||
| $|x-3|<1$ | ||||
| Then $|x+3|<7$ | ||||
| $\delta=\min(1,\varepsilon/7)$ |
Match each proof move to the reason it is needed.
| Keeps $x$ in a bounded interval | Turns the variable factor into a fixed bound | Separates the controllable small factor | Satisfies both the neighborhood and accuracy constraints | |
|---|---|---|---|---|
| Require $|x-2|<1$ | ||||
| Then $|x+2|<5$ | ||||
| $|x^2-4|=|x-2||x+2|$ | ||||
| Choose $\delta=\min(1,\varepsilon/5)$ |
Match each step in a proof of $\lim_{x\to a}x^2=a^2$ to the reason it is needed.
| Separates the factor controlled by delta | Places $x$ in a bounded neighbourhood of $a$ | Replaces the varying factor by a fixed constant | Satisfies both the neighbourhood restriction and epsilon bound | |
|---|---|---|---|---|
| Write $|x^2-a^2|=|x-a||x+a|$ | ||||
| Require $|x-a|<1$ | ||||
| Conclude $|x+a|<2|a|+1$ | ||||
| Choose $\delta=\min(1,\varepsilon/(2|a|+1))$ |
Proving $\lim_{x \to 5} x^2 = 25$: after restricting to $|x - 5| < 1$, what is the bound on $|x + 5|$?
Answer:
Proving $\lim_{x \to 2} x^2 = 4$ with $\varepsilon = 2/10$: if the restriction is $|x - 2| < 1$, which $\delta$ does the standard construction give?
Answer:
$|f(x) - L| \le 2\,|x - a|$ and $|g(x) - M| \le 2\,|x - a|$. To force $|(f + g)(x) - (L + M)| < 8/10$, how small must $|x - a|$ be?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A proof ends $\delta = \min\bigl(1, \varepsilon/3\bigr)$. Why the $\min$?
You can construct a $\delta$ for a quadratic, a cubic and a reciprocal limit, and say what each half of the minimum protects. Say in your own words what goes wrong if the $\min$ is dropped. Next: the limit laws, proved once so the definition need not be used again.
8. Your turn: find $\delta$ for $\lim_{x \to 4} x^2 = 16$, step 3
Divide, and take the min.