Exercises — Limit laws — sum, product, quotient, constant multiple

• $\underset{x\to 5}{\lim }7===7==$. This is the constant atom ${\lim }_{x\to a}c=c$: a function that is always $7$ heads to $7$ no matter where $x$ goes.
• $\underset{x\to 5}{\lim }x===5==$. This is the identity atom ${\lim }_{x\to a}x=a$: the function that just returns its input heads to $a=5$. These two are the only facts we are allowed to assume — every other limit is built from them.

Call $L=4$ and $M=-2$.

• (a) Sum law: $L+M=4+(-2)===2==$.
• (b) Constant-multiple law: $3L=3\cdot 4===12==$.
• (c) Product law: $LM=4\cdot (-2)===-8==$.
• (d) Quotient law — first check the bottom: $M=-2\ne 0$, so it is legal. $\frac{L}{M}=\frac{4}{-2}===-2==$. Why check (d)'s denominator? The quotient law carries a fine-print clause $M\ne 0$. Here $M=-2$, so we pass the bouncer.

Split with the sum and constant-multiple laws: $=4{\lim }_{x\to 2}{x}^3-{\lim }_{x\to 2}x+{\lim }_{x\to 2}5.$ The power law ($\lim x^n=a^n$, which is the product law applied $n$ times) gives $\lim x^3=2^3=8$. $=4(8)-2+5=32-2+5===35==.$ Because this is a polynomial, the whole detour equals $p(2)$ — direct substitution works.

Check the bottom first: ${\lim }_{x\to 1}(x+4)=1+4=5\ne 0$. Legal.

• Top: ${\lim }_{x\to 1}(x^2+3x)=1+3=4$.
• Bottom: $5$. Quotient law: $\frac{4}{5}===0.8==$.

Inside: ${\lim }_{x\to 3}(x^2+7)=9+7=16\ge 0$, so the root law applies. $=\sqrt{16}===4==.$ Why does the root law hold? $\sqrt{\cdot }$ is continuous, so limits pass through it — see Continuity.

Check the bottom: ${\lim }_{x\to 4}(x-4)=0$. The quotient law is forbidden — this is an $0/0$ form (top also $\to 0$). Factor the top as a difference of squares: $\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4(x\ne 4).$ Why is cancelling legal? The limit only watches $x$ near $4$, never $x=4$ itself, so $x-4\ne 0$ there — see the figure: the graph is the line $y=x+4$ with a single hole punched at $x=4$.

Now apply the sum law: ${\lim }_{x\to 4}(x+4)=4+4===8==$.

Bottom $\to 0$, top $\to \sqrt{9}-3=0$ — another $0/0$. We cannot factor nicely, so we rationalize by multiplying by the conjugate $\sqrt{x+9}+3$ over itself (this is multiplying by $1$, so nothing changes): $\frac{\sqrt{x+9}-3}{x}\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{x}{x(\sqrt{x+9}+3)}.$ Why rationalize? It turns the stubborn $\sqrt{}-3$ into a plain $x$ on top that cancels the $x$ below. For $x\ne 0$ cancel the $x$: $=\frac{1}{\sqrt{x+9}+3}.$ Now the bottom's limit is $\sqrt{9}+3=6\ne 0$, so the quotient law is finally legal: $=\frac{1}{6}\approx ==0.16667==.$

Bottom: $\lim (x^2-4)=4-4=0$ — quotient law forbidden. Check the top: $\lim (x^2+5x+6)=4-10+6=0$. So it's $0/0$; factor both. $\frac{x^2+5x+6}{x^2-4}=\frac{(x+2)(x+3)}{(x+2)(x-2)}=\frac{x+3}{x-2}(x\ne -2).$ Why did both share $(x+2)$? Because $x=-2$ is a root of both — that shared factor is exactly what causes the $0/0$. Now the new bottom's limit is $-2-2=-4\ne 0$, so the quotient law applies: $=\frac{-2+3}{-2-2}=\frac{1}{-4}===-0.25==.$

Split into One-sided limits:

• For $x>0$, $|x|=x$, so $f(x)=\frac{x}{x}=1$. Thus the right limit is ${\lim }_{x\to 0^{+}}f===1==$.
• For $x<0$, $|x|=-x$, so $f(x)=\frac{-x}{x}=-1$. Thus the left limit is ${\lim }_{x\to 0^{-}}f===-1==$.

The two sides disagree ($1\ne -1$), so the two-sided limit does not exist. Which law does this warn us about? The quotient law needed both $\lim$(top) and $\lim$(bottom) to exist first. Here ${\lim }_{x\to 0}x=0$ makes the bottom head to $0$ anyway, and the assembled function has no single limit — a live reminder that the laws only combine limits that already exist.

The product law only runs forward: it needs both parts' limits to exist first. Running it backward is false. Counterexample at $a=0$: let $f(x)=\{\begin{array}{ll}1& x\ge 0\\ -1& x<0\end{array},g(x)=f(x).$ Neither ${\lim }_{x\to 0}f$ nor ${\lim }_{x\to 0}g$ exists (left value $-1$, right value $+1$ — same jump as Exercise 4.2). But their product is $f(x)\cdot g(x)=f(x{)}^2=(\pm 1{)}^2=1\text{ for all }x,$ so $\underset{x\to 0}{\lim }(fg)===1==$ exists. The product's limit existing tells us nothing about the pieces. $■$

This is the Derivative as a limit — and it is a $0/0$ form in disguise, so the quotient law is initially blocked. $\frac{h(x)-h(3)}{x-3}=\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3),$ using difference of squares to factor and then cancelling (legal because $x\ne 3$ near the limit point). Now the sum law finishes it: $h^{\prime }(3)={\lim }_{x\to 3}(x+3)=3+3===6==.$ Notice the pattern: every derivative-from-definition is a $0/0$ quotient that the limit laws rescue by cancelling the offending $(x-a)$. That is why the sum/product/quotient rules of differentiation all descend from these four limit laws.

Which law has a side-condition, and what is it? ::: The quotient law; needs $M\ne 0$. First reflex on any quotient limit? ::: Evaluate the denominator's limit before anything else. What does $0/0$ actually mean? ::: Indeterminate — the plain law is silent; factor/cancel/rationalize. Rescue tool for  ​-type $0/0$? ::: Multiply by the conjugate (rationalize). Do the laws run backwards? ::: No — they need the pieces' limits to exist first. Why is every derivative a limit-law problem? ::: It's a $0/0$ quotient rescued by cancelling $(x-a)$.