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Evaluate limits by factoring, conjugates, standard limits and leading terms; the three conditions for continuity, the kinds of discontinuity, and the intermediate value theorem.
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 can evaluate a limit algebraically rather than by reading a graph: substituting first, and — when that gives $\tfrac{0}{0}$ — factoring, using a conjugate, or matching a standard limit until substitution is safe. You can also say precisely what continuity requires, name a discontinuity by which condition it breaks, choose a constant that joins two pieces of a function, and use the intermediate value theorem for what it actually promises.
You have met limits informally: what a function approaches, read off a graph or a table. This lesson makes them algebraic. The one new habit is that $\tfrac{0}{0}$ is not an answer — it is an instruction to rewrite the expression until substitution becomes safe.
Limit: the value $f(x)$ approaches as $x$ approaches $a$ — never what $f(a)$ is.
Indeterminate form: $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$, which say nothing until the expression is rewritten.
One-sided limit: the value approached from the left ($x \to a^-$) or from the right ($x \to a^+$).
Conjugate: $\sqrt{A} + B$ paired with $\sqrt{A} - B$; multiplying them clears the root.
$\lim_{x \to a} f(x) = L$ says that $f(x)$ can be made as close to $L$ as you like by taking $x$ close enough to $a$ — without ever using $x = a$ itself. That exclusion is what makes limits useful: it lets us talk about a function's behaviour at a point where it is undefined, which is exactly where derivatives live.
In practice there are four moves and a rule for choosing between them.
Substitute first. If it gives a number, that is the limit and there is nothing more to do.
If it gives $\tfrac{0}{0}$, rewrite. Factor and cancel when the expression is polynomial; multiply by the conjugate when a square root is in the way; use the standard limit $\lim_{x \to 0} \tfrac{\sin x}{x} = 1$ when a sine is. Each of these produces a new expression agreeing with the old one everywhere except at the point — which is all the limit sees.
At infinity, compare leading terms. Divide by the highest power in the denominator; everything else vanishes. The answer is the ratio of the leading coefficients when the degrees match, $0$ when the bottom wins, and infinite when the top does.
Check both sides when the function is piecewise or has an absolute value. A limit exists only when the left and right limits both exist and agree.
Another way: picture
Picture a graph with a hole punched in it at $x = a$. Walk along the curve from the left and from the right: both walks head for the same height, and that height is the limit. Whether the point at $a$ is missing, or drawn somewhere else entirely, changes nothing about either walk.
Another way: steps
| What substitution gives | What it means | The move |
|---|---|---|
| A number | The function is continuous there | Done |
| $\tfrac{k}{0}$, $k \ne 0$ | A vertical asymptote | Check the sign on each side |
| $\tfrac{0}{0}$, polynomial | A common factor | Factor and cancel |
| $\tfrac{0}{0}$, with a root | A hidden common factor | Multiply by the conjugate |
| $\tfrac{0}{0}$, with a sine | The standard limit | Match $\tfrac{\sin u}{u}$ |
| $\tfrac{\infty}{\infty}$ | Leading terms decide | Divide by the highest power |
The middle rows are the whole of the technique. $\tfrac{0}{0}$ never means 'no answer'; it means the expression is written in a form that hides one.
"$\tfrac{0}{0}$ means the limit is zero, or one, or does not exist." It means nothing at all yet. Two different functions of that form can have any two limits you like; the form only tells you to keep working.
"The limit is $f(a)$." For a continuous function it happens to be, which is the whole point of continuity — but a limit is defined without ever evaluating $f$ at $a$, and the interesting cases are the ones where $f(a)$ is missing or different.
"A limit at infinity is a number the function reaches." It is a value the function approaches and generally never attains. $\tfrac{1}{x} \to 0$ and is never zero.
Substituting gives $\tfrac{0}{0}$, so a factor cancels.
Not an answer — an instruction.
$x^2 - x - 6 = (x - 3)(x + 2)$, so the expression is $x + 2$ for every $x \ne 3$.
Factor and cancel.
The limit is $3 + 2 = 5$.
Substitute into the new expression.
Divide every term by $x^2$: $\dfrac{4 - 1/x}{3 + 7/x^2}$.
Highest power below.
$1/x$ and $7/x^2$ both vanish.
What is left is what matters.
The limit is $\tfrac{4}{3}$ — which is also the horizontal asymptote of the graph.
Substituting gives $\tfrac{0}{0}$, and there is a root, so multiply by the conjugate $\sqrt{x + 5} + 3$.
The numerator becomes $(x + 5) - 9 = x - 4$, which cancels the denominator.
What is left is $\dfrac{1}{\sqrt{x+5}+3}$, which at $x = 4$ is $\tfrac{1}{6}$.
Find $\displaystyle\lim_{x \to 9} \frac{x^2 - 14x + 45}{x - 9}$.
answer
Find $\displaystyle\lim_{x \to 2} \frac{\sqrt{x + 14} - 4}{x - 2}$.
answer
Find $\displaystyle\lim_{x \to \infty} \frac{6x^2 + 7x}{3x^2 + 7}$.
answer
Find $\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{9x}$.
answer
You can evaluate a limit algebraically, and you know that a limit never looks at the point itself. Continuity is the condition that says the point and the limit agree — the case where all the machinery of the last section is unnecessary.
Continuous at $a$: $f(a)$ exists, $\lim_{x \to a} f(x)$ exists, and they are equal.
Removable discontinuity: the limit exists but the value is missing or different — a hole.
Jump discontinuity: the one-sided limits exist and disagree.
Infinite discontinuity: a vertical asymptote.
Intermediate value theorem: a continuous function on $[a, b]$ takes every value between $f(a)$ and $f(b)$.
$f$ is continuous at $a$ when three separate things hold:
Each can fail on its own, and which one fails names the discontinuity. If only the third fails — the limit exists but the value is wrong or missing — the discontinuity is removable: redefining $f$ at that one point repairs it. If the one-sided limits exist but disagree, it is a jump, and no redefinition helps. If the values grow without bound, it is infinite.
Polynomials are continuous everywhere; rational functions everywhere their denominator is not zero; $\sin$, $\cos$ and $e^x$ everywhere; $\ln x$ on $x > 0$; and sums, products, quotients and composites of continuous functions are continuous wherever they are defined. So in practice the only points worth checking are where a denominator vanishes and where a piecewise definition changes.
The intermediate value theorem is what continuity buys: if $f$ is continuous on the closed interval $[a, b]$ and $N$ lies between $f(a)$ and $f(b)$, then $f(c) = N$ for some $c$ in $(a, b)$. It is how a sign change is turned into a root — and it is an existence statement, saying nothing about where the root is or how many there are.
Another way: picture
Picture three broken graphs at the same $x$: one with a hole and a stray dot floating above it, one that stops at one height and restarts at another, and one that shoots off to infinity. The first can be mended by moving a single dot. The other two cannot be mended at all.
Another way: steps
"Continuous means you can draw it without lifting your pen." A useful picture, and not the definition. It is about three conditions at each point, and there are continuous functions no hand could draw.
"If $f(a)$ exists, $f$ is continuous at $a$." The value existing is one condition out of three. A jump function is defined at the jump and is not continuous there.
"The intermediate value theorem finds the root." It says one exists. Nothing in it locates a root, and it says nothing at all about how many there are.
$f(x) = 3x + 1$ for $x \le 2$ and $x^2 + k$ for $x > 2$.
One join, at $x = 2$.
From the left: $3(2) + 1 = 7$. From the right: $4 + k$.
Both sides.
$4 + k = 7$, so $k = 3$ — and no other value works.
$f(x) = x^3 - x - 1$ is a polynomial, so continuous on $[1, 2]$.
Continuity is the hypothesis.
$f(1) = -1$ and $f(2) = 5$, so zero lies between them.
A sign change.
So $f$ has a root in $(1, 2)$ — though the theorem gives no hint where, and finding it needs quite different tools.
$f(2)$ is $\tfrac{0}{0}$ — undefined, so condition 1 already fails.
The limit does exist: cancelling gives $x + 2$, which tends to $4$.
So it is a removable discontinuity, and defining $f(2) = 4$ would make the function continuous everywhere.
A function has $\lim_{x \to 2} f(x) = 4$ and $f(2) = 5$. Is it continuous at $x = 2$?
$f(x) = 6x + 7$ for $x \le 6$ and $f(x) = x^2 + c$ for $x > 6$. What value of $c$ makes $f$ continuous at $x = 6$?
answer
$f$ is continuous on $[\,2, 6\,]$ with $f(2) = -8$ and $f(6) = 8$. What follows?
An electricity tariff charges $2$ pence a unit for the first $25$ units and $c$ pence a unit thereafter, plus a standing charge of $7$ pence that applies only above $25$ units. For the total bill to have no jump at $25$ units, what must $c$ be?
answer
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = 5x$ for $x < 8$ and $f(x) = x + 40$ for $x \ge 8$. Does $\lim_{x \to 8} f(x)$ exist?
Match each behaviour to the kind of discontinuity it is.
| Removable | Jump | Infinite | |
|---|---|---|---|
| A hole; the limit exists but the value does not match | |||
| The two one-sided limits disagree | |||
| A vertical asymptote |
You can evaluate limits and reason about continuity. Without looking: what does $\tfrac{0}{0}$ tell you, and what are the three conditions for continuity at a point?
9. Your turn: $\displaystyle\lim_{x \to 4} \frac{\sqrt{x + 5} - 3}{x - 4}$, step 3
20. Your turn: is $f(x) = \dfrac{x^2 - 4}{x - 2}$ continuous at $x = 2$?, step 3