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Proof by induction — base case, hypothesis, step and conclusion — and limits by substitution, by cancelling, from one side, and at infinity.
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.
This lesson holds the two ideas that make calculus possible. The first is proof by induction: a way of establishing a statement for every integer by checking one case and building a link from each case to the next, which is the only honest answer to a claim about infinitely many numbers. The second is the limit — what a function is heading towards near a point, whether or not it is even defined there. You will evaluate limits by substituting, by cancelling away an indeterminate form, by taking one side at a time, and by comparing leading terms far out.
You can substitute a number into a formula and check whether an equation is true, and you have just used the formula $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$. You have probably also noticed that no amount of checking cases proves a statement about all integers. Induction is the argument that closes that gap.
$P(n)$: the statement, for a particular $n$.
Base case: the first value, checked outright.
Inductive hypothesis: the assumption that $P(k)$ holds for one unnamed $k$.
Inductive step: the proof that $P(k)$ implies $P(k+1)$.
Conclusion: therefore $P(n)$ for every $n$ from the base case on.
Counterexample: a single case that fails, which is all it takes to disprove a claim.
To prove a statement $P(n)$ for every integer from some starting point on, you prove two things. First the base case: $P(1)$, checked by direct calculation. Then the inductive step: if $P(k)$ holds for some $k$, then $P(k+1)$ holds. Together these force $P(n)$ for every $n$, because $P(1)$ gives $P(2)$, which gives $P(3)$, and any particular $n$ is finitely many steps along. For $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$: the base case is $1 = \frac{1 \times 2}{2}$, true. For the step, assume the sum to $k$ is $\frac{k(k+1)}{2}$; the sum to $k+1$ is then $\frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$, which is the formula with $k+1$ in it. Done. Two warnings. The base case is not a formality: without it the chain has nothing to start from. And the step must move by one; a step from $k$ to $k+2$ proves the statement only for the numbers that chain actually lands on, which is every other one.
Another way: picture
A row of dominoes standing up, numbered $1, 2, 3, \ldots$. A hand tips the first one over (the base case), and each domino is close enough to knock over the next (the step). Remove either and the row stays standing.
Another way: steps
"Assuming $P(k)$ is circular — you assumed what you are proving." No. You assume it for one value and prove the link to the next. What is proved outright is a conditional, and the base case is what makes the chain start.
Skipping the base case. The step alone proves nothing. "If $n = n + 1$ then $n + 1 = n + 2$" is a perfectly valid step for a statement that is never true.
Not writing down the target. The step has to reach the formula with $k+1$ substituted in. Writing that target out first turns the step from a fishing expedition into a calculation with a destination.
Base case: $2^1 = 2 > 1$. True.
Arithmetic, not algebra.
Assume $2^k > k$. Then $2^{k+1} = 2 \times 2^k > 2k$.
Use the assumption exactly once.
And $2k = k + k \ge k + 1$ whenever $k \ge 1$, so $2^{k+1} > k+1$: the target reached.
The chain reaches $1$, then $3$, then $5$, and so on.
Follow it and see where it lands.
Nothing at all has been said about $2$, $4$ or $6$, so the claim 'for all $n$' is unproved — though the odd case is genuinely proved.
Left side at $n = 2$: $3^2 = 9$.
Right side: $2 \times 2 + 1 = 5$. Since $9 > 5$, the base case holds.
Put the four parts of a proof by induction, that a statement $P(n)$ holds for every $n \ge 3$, into order.
Number the steps in order (write the number in the box):
You are proving $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$ by induction and have assumed it for $n = 9$. What number must the sum $1 + 2 + \cdots + 10$ come out as?
Answer:
To prove $3^n > n + 2$ for all $n \ge 2$, the base case checks $n = 2$. Give the two sides.
left = l, right = r
A proof checks $P(1)$ and then shows that $P(k)$ implies $P(k + 2)$. Which $n$ does it prove $P(n)$ for?
You can substitute a number into a formula, cancel a common factor, and compare the degrees of a rational function's top and bottom — which you did when you found horizontal asymptotes earlier in this course. A limit is that asymptote idea made local: instead of asking where the graph goes far away, you ask where it goes near a particular point.
Limit: the value $f(x)$ closes in on as $x$ closes in on a point, written $\lim_{x \to c} f(x)$.
One-sided limit: approaching only from below ($x \to c^-$) or only from above ($x \to c^+$).
Continuous at $c$: no break there, so the limit equals $f(c)$.
Indeterminate form: $\frac{0}{0}$ — a signal to do algebra, not an answer.
Does not exist: no single number is being approached.
At infinity: what the function settles on as $x$ grows without bound.
$\lim_{x \to c} f(x) = L$ means: $f(x)$ can be made as close to $L$ as you like by taking $x$ close enough to $c$ — without ever putting $x = c$. That last clause is the whole point. Three techniques cover almost everything at this level. Substitute when the function is continuous: $\lim_{x\to 2}(3x+1) = 7$, and the same for any polynomial or root that is defined there, so $\lim_{x\to2}\sqrt{x+7} = 3$. Cancel when substitution gives $\frac{0}{0}$: $\frac{x^2-4}{x-2}$ equals $x+2$ everywhere except at $2$, and since the limit ignores the point itself, the answer is $4$. Compare leading terms at infinity: $\lim_{x\to\infty}\frac{2x+1}{x+5} = 2$, the ratio of the leading coefficients — the same rule that gave horizontal asymptotes. And a limit can fail to exist. For a one-sided question use the piece of the definition that applies on that side; if the two sides give different answers, as for $\frac{1}{x}$ at $0$, then no two-sided limit exists.
Another way: picture
The line $y = x + 2$ with a small hollow circle at $(2, 4)$. Arrows creep along the line towards the hole from both sides; the height they approach is $4$, whatever is or is not drawn at the hole itself.
Another way: steps
"If $f(c)$ is undefined, the limit does not exist." The limit never looks at the point itself. $\frac{x^2-4}{x-2}$ is undefined at $2$ and its limit there is $4$, which is precisely why limits are worth having.
"$\frac{0}{0} = 0$" or "$= 1$". It is neither; it is a form that carries no information. Cancel, rationalise or factor until the form goes away, and only then substitute.
Ignoring one side. For $\frac{1}{x}$ at $0$ the right side runs to $+\infty$ and the left to $-\infty$. A limit exists only if both sides agree, so this one does not exist at all.
Substituting gives $\frac{0}{0}$, so there is a common factor to remove.
An indeterminate form is an instruction, not an answer.
$\frac{(x-3)(x+3)}{x-3} = x + 3$ for $x \ne 3$, and the limit ignores $x = 3$, so the answer is $6$.
Divide top and bottom by $x^2$: $\dfrac{3 + 1/x^2}{1 - 4/x^2}$.
Highest power on both.
The small terms vanish, leaving $\frac{3}{1} = 3$ — the ratio of the leading coefficients.
The root is defined and unbroken near $x = 2$, so substitute.
$\sqrt{2 + 7} = \sqrt{9} = 3$.
What is $\displaystyle\lim_{x \to 1} \left(7x + 2\right)$?
Answer:
What is $\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$?
Answer:
What is $\displaystyle\lim_{x \to \infty} \dfrac{6x + 5}{7x + 3}$?
Answer:
$f(x) = x + 1$ for $x < 3$, and $f(x) = 12$ for $x \ge 3$. What is $\displaystyle\lim_{x \to 3^-} f(x)$?
Answer:
A stone has fallen $9t^2$ metres after $t$ seconds. Over the interval from $t = 1$ to $t = 1 + h$ its average speed works out as $9(2 \times 1 + h)$ metres per second. What speed is the stone travelling at the instant $t = 1$, in metres per second?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Why does a completed induction prove a statement for every positive integer, when only one case was actually checked?
What is $\displaystyle\lim_{x \to 0} \dfrac{5}{x}$?
You can build an induction and evaluate a limit. Without looking: why is the base case not a formality, and why can $\lim_{x \to 2}\frac{x^2-4}{x-2}$ be $4$ when the function has no value at $2$?
8. Your turn: the base case of $3^n > 2n + 1$ for $n \ge 2$, step 2
19. Your turn: $\lim_{x \to 2} \sqrt{x + 7}$, step 2