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Sum, scalar, product, quotient and power; why substitution works for polynomials; and what a law that does not apply tells you.
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 combine known limits with the sum, scalar, product, quotient and power laws, name each law as you use it and check the hypothesis it carries, explain why every polynomial limit is settled by substitution and why a rational one is only settled where the denominator survives, recognise that a law whose hypothesis fails has given you no answer rather than the answer $0/0$, and say why the laws never run backwards.
Suppose $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$ — both existing is the hypothesis every line below shares. Then:
| Law | Statement | Extra hypothesis |
|---|---|---|
| sum | $\lim (f + g) = L + M$ | — |
| scalar | $\lim (cf) = cL$ | — |
| product | $\lim (fg) = LM$ | — |
| quotient | $\lim (f/g) = L/M$ | $M \ne 0$ |
| power | $\lim f^n = L^n$ | $n$ a positive integer |
| root | $\lim \sqrt[n]{f} = \sqrt[n]{L}$ | $L > 0$ for even $n$ |
Each is proved once from the epsilon-delta definition, and after that the definition is never used for a routine limit again. The sum law is the $\varepsilon/2$ argument of the last lesson; the product law is an $\varepsilon/2$ argument with a boundedness step; the quotient law is the reciprocal construction with the restriction that keeps the denominator from zero.
A theorem is its hypotheses. Quoting the conclusion of one whose hypotheses you have not checked is not a proof of anything — and in this course it is the single most common way a correct-looking argument turns out to be wrong.
The two hypotheses that actually bite:
$M \ne 0$ in the quotient law. When $M = 0$ the law is silent. It does not say the limit fails to exist; it says nothing, and something else has to settle the question. That is exactly the $0/0$ case, and it is the interesting one.
Both limits existing. The laws build limits out of limits. They never run backwards: $\lim (f + g)$ existing tells you nothing about $\lim f$.
Another way: picture
The laws as a set of pipes: limits go in, a limit comes out. Each pipe has a gate on it — the hypothesis — and a gate that is shut does not divert the flow somewhere else, it stops it. The quotient pipe's gate is closed exactly when $M = 0$, and that is when you have to go round by hand.
Another way: steps
It is worth seeing that the everyday shortcut is a theorem and not a convention.
Two limits are proved directly from the definition, and they are the only two anyone proves: $\lim_{x \to a} c = c$ (take any $\delta$) and $\lim_{x \to a} x = a$ (take $\delta = \varepsilon$).
From those, the product law gives $\lim x^2 = a^2$, and again $\lim x^n = a^n$. The scalar law gives $\lim c x^n = c a^n$. The sum law adds the terms. So for any polynomial $p$, $$\lim_{x \to a} p(x) = p(a),$$ and for a rational function $p/q$ the quotient law adds the same conclusion wherever $q(a) \ne 0$.
That is the whole justification for substituting, and it comes with its exception attached: at a point where the denominator vanishes, substitution is not permitted, which is precisely where the interesting limits live.
Quoting the quotient law at $0/0$ and concluding. The law is silent there. Writing "$= 0/0$, so no limit" quotes a law that declined to answer as though it had answered.
Running a law backwards. $\lim(f+g)$, $\lim(fg)$ or $\lim(f/g)$ existing implies nothing about the pieces. $f = |x|/x$ and $g = -|x|/x$ settles all three at once.
Using the power law with a non-integer exponent. $\lim f^{1/2} = L^{1/2}$ needs $L > 0$ and the continuity of the square root, which is a later theorem, not this law.
Splitting a limit that is not a sum of limits. $\lim (f \cdot g)$ where $f \to 0$ and $g \to \infty$ is not $0$; the product law does not apply, because $\lim g$ does not exist.
The commonest error in this lesson is not arithmetic. It is treating "$0/0$" as something the quotient law said, and then reporting it. The law has a hypothesis; when the hypothesis fails the law contributes nothing to the discussion, and the symbols $0/0$ are a note about why, not a result.
The same habit, further on, produces "$\infty - \infty = 0$" and "$0 \times \infty = 0$". Each is a law quoted outside its hypotheses. The discipline that prevents all of them is one sentence long: name the law, check the hypothesis, then use it.
$\lim_{x \to 2} \dfrac{3x^2 - x}{x + 4}$. The denominator tends to $6 \ne 0$, so the quotient law applies.
Check the hypothesis first.
Numerator: $3 \cdot 4 - 2 = 10$ by the power, scalar and sum laws.
Every step named.
So the limit is $\tfrac{10}{6} = \tfrac53$ — which is $p(2)/q(2)$, exactly as substitution would give, and now with a reason.
$\lim_{x \to 1} \dfrac{x^2 - 1}{x^2 - 3x + 2}$. Both numerator and denominator tend to $0$, so the quotient law is silent.
Stop. Do not conclude.
Factor: $\dfrac{(x-1)(x+1)}{(x-1)(x-2)} = \dfrac{x+1}{x-2}$ for $x \ne 1$.
Algebra takes over.
Now the denominator tends to $-1 \ne 0$, the quotient law applies, and the limit is $\tfrac{2}{-1} = -2$.
The law resumes once its hypothesis is true.
Both tend to $0$, so the quotient law says nothing. Factor both.
Check, then stop.
$\dfrac{(x-3)(x+3)}{(x-3)(x+1)} = \dfrac{x+3}{x+1}$ for $x \ne 3$.
Denominator now tends to $4 \ne 0$: the limit is $\tfrac{6}{4} = \tfrac32$.
Match each limit form to the valid next step.
| Use direct substitution | Apply the quotient law | Simplify or use another justified method | Analyze one-sided divergence, not an indeterminate form | |
|---|---|---|---|---|
| A polynomial at a finite input | ||||
| A quotient with nonzero denominator limit | ||||
| A quotient producing $0/0$ | ||||
| A quotient producing $5/0$ |
Match each operation to the hypothesis that makes its limit rule valid.
| Both component limits exist | $\lim g(x)\ne0$ | Use continuity of the polynomial | The limiting value of $f$ is nonzero | |
|---|---|---|---|---|
| $\lim(f+g)$ | ||||
| $\lim f/g$ | ||||
| $\lim p(f(x))$ for polynomial $p$ | ||||
| $\lim 1/f(x)$ |
Match each limit-law situation to the conclusion or next move it supports.
| The sum law gives $\lim(f+g)=5$ | The quotient law gives $\lim(f/g)=-7/2$ | Simplify or analyze further; the quotient law does not apply | The composition law gives $\lim f(g(x))=f(4)$ | |
|---|---|---|---|---|
| $\lim f=2$ and $\lim g=3$ | ||||
| $\lim f=7$ and $\lim g=-2$ | ||||
| $\lim f=0$ and $\lim g=0$ in $f/g$ | ||||
| $\lim g(x)=4$ and $f$ is continuous at 4 |
$\lim_{x \to a} f(x) = 5$ and $\lim_{x \to a} g(x) = 5$. Find $\lim_{x \to a} (3f(x) + g(x))$.
Answer:
$\lim_{x \to a} f(x) = 3$ and $\lim_{x \to a} g(x) = 2$. Find $\lim_{x \to a} \dfrac{f(x)}{g(x)}$.
Answer:
Find $\displaystyle\lim_{x \to 5} \frac{1}{x + 1} \cdot \frac{x^2 - 25}{x - 5}$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A student writes: "$\lim_{x \to 2} \dfrac{x^2 - 4}{x - 2} = \dfrac{\lim (x^2 - 4)}{\lim (x - 2)} = \dfrac{0}{0}$, so the limit does not exist." What is wrong?
You can combine limits with the laws and say which hypothesis each needs. Say in your own words what the quotient law tells you when the denominator's limit is zero. Next: the squeeze theorem, which settles limits the laws cannot reach at all.
8. Your turn: $\lim_{x \to 3} \dfrac{x^2 - 9}{x^2 - 2x - 3}$, step 3