Chapter 12 · Limits and Derivatives

12.1 Intuitive Idea of Limits

$\displaystyle\lim_{x\to a}f(x)=l$ says that $f(x)$ gets as close to $l$ as we like whenever $x$ is close enough to $a$ — from both sides, and whatever happens at $x=a$ itself. Learn to read limits from tables and graphs, split them into left-hand and right-hand limits, and combine them with the algebra of limits.

A stone is dropped from the top of a tall building. Physics tells us that after $t$ seconds it has fallen about $s=4.9t^2$ metres. How fast is it moving at the instant $t=2\ \text{s}$? Average speed is easy: between $t=2$ and $t=3$ the stone falls $4.9\times9-4.9\times4=24.5\ \text{m}$ in $1\ \text{s}$, an average of $24.5\ \text{m/s}$. But the stone speeds up all the time, so that average mixes slow and fast moments. A shorter interval gives a better answer — and the shorter the better. What we cannot do is take an interval of length zero: then the stone falls $0\ \text{m}$ in $0\ \text{s}$ and the "average" is $\tfrac00$, which has no value at all.

The way out is the central idea of this chapter. We calculate the average speed over intervals of length $h$, let $h$ get closer and closer to $0$ without ever being $0$, and ask what number the averages approach. That number is called a limit. In this section we make the idea precise enough to work with: we read limits from tables and graphs, see why the value of a function at a point does not affect its limit there, split a limit into a left-hand and a right-hand limit, and learn the rules that let us combine limits.

From Average Speed to Instantaneous Speed

For the falling stone, the average speed between $t=2$ and $t=2+h$ (where $h$ may be positive or negative, but not $0$) is $$\text{average speed}=\dfrac{s(2+h)-s(2)}{h}=\dfrac{4.9(2+h)^2-4.9\times4}{h}.$$ Expanding the bracket, $4.9(4+4h+h^2)-19.6=19.6h+4.9h^2$, so for every $h\ne0$ $$\text{average speed}=\dfrac{19.6h+4.9h^2}{h}=19.6+4.9h\ \text{m/s}.$$ We were allowed to cancel $h$ because $h\ne0$. Now make $h$ small:

$h$ (s)	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
average speed (m/s)	$19.11$	$19.551$	$19.5951$	$19.6049$	$19.649$	$20.09$

The averages crowd in on $19.6$ from both sides: intervals ending just before $t=2$ give slightly less, intervals starting at $t=2$ give slightly more, and the error is exactly $4.9h$, which can be made as small as we please. So we say that the instantaneous speed of the stone at $t=2$ is $19.6\ \text{m/s}$ (about $70.6\ \text{km/h}$). In symbols, $$\lim_{h\to0}\dfrac{s(2+h)-s(2)}{h}=19.6,$$ read "the limit, as $h$ tends to $0$, of … is $19.6$".

The same number has a picture. On the graph of $s$ against $t$, the average speed $\tfrac{s(2+h)-s(2)}{h}$ is the slope (Section 9.1) of the chord joining the points at $t=2$ and $t=2+h$. As $h\to0$ the chord swings round towards the line that just touches the curve at $t=2$ — the tangent — so $19.6$ is the slope of the curve at that point. Instantaneous speed and the slope of a tangent are the same idea, the derivative, which we study properly in Section 12.3. First we need the language of limits.

✏️ Example 1 — Average speed over shrinking intervals

A car moving along a straight road is $s=t^2+4t$ metres from a signal $t$ seconds after it starts. (a) Find the average speed between $t=3$ and $t=3+h$, $h\ne0$. (b) Use a table to estimate the speed at $t=3$. (c) Show that the error of the average is exactly $|h|$.

💡 Show solution▶

(a) $s(3)=9+12=21$ and $s(3+h)=(3+h)^2+4(3+h)$ $=9+6h+h^2+12+4h$ $=21+10h+h^2$. So the average speed is $\dfrac{10h+h^2}{h}=10+h\ \text{m/s}$ for $h\ne0$.

(b) $h=0.1,0.01,0.001$ give $10.1,\ 10.01,\ 10.001$; $h=-0.1,-0.01,-0.001$ give $9.9,\ 9.99,\ 9.999$. The averages approach $10$, so the speed at $t=3$ is $10\ \text{m/s}$ ($36\ \text{km/h}$).

Check Directly: $s(3.001)-s(3)$ $=(9.006001+12.004)-21$ $=0.010001$, and $\tfrac{0.010001}{0.001}=10.001$ ✓.

The Limit of a Function

Look again at what we did. We had an expression in $h$, we kept $h\ne0$, and we watched the values as $h$ got close to $0$. The same idea works for any function $f$ and any point $a$.

📘 Definition 12.1 — Limit of a function

Let $f$ be defined at every point of some open interval containing $a$, except possibly at $a$ itself. We say that the limit of $f(x)$ as $x$ tends to $a$ is $l$, and write $$\lim_{x\to a}f(x)=l,$$ if the values $f(x)$ can be made as close to the number $l$ as we like by taking $x$ close enough to $a$, on either side of $a$, with $x\ne a$. We also write $f(x)\to l$ as $x\to a$.

Two words in the definition carry all the weight. "Either side": $x$ may approach $a$ from below or from above, and both approaches must lead to the same $l$. "$x\ne a$": the limit describes what $f$ does near $a$, not at $a$. The value $f(a)$ may be undefined, or equal to $l$, or something else entirely — it makes no difference to the limit.

✏️ Example 2 — A limit at a hole

Let $f(x)=\dfrac{x^2-4}{x-2}$. (a) Is $f(2)$ defined? (b) Make a table of $f(x)$ for $x$ near $2$ and estimate $\displaystyle\lim_{x\to2}f(x)$. (c) Describe the graph of $f$.

💡 Show solution▶

(a) $f(2)=\tfrac{0}{0}$, which is not a number: $2$ is not in the domain of $f$.

(b) $x=1.9,\ 1.99,\ 1.999$ give $f(x)=3.9,\ 3.99,\ 3.999$; $x=2.1,\ 2.01,\ 2.001$ give $4.1,\ 4.01,\ 4.001$. From both sides the values approach $4$, so $\displaystyle\lim_{x\to2}f(x)=4$.

Why For $x\ne2$, $\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$. Near $2$ (but not at $2$) $f(x)=x+2$, which is close to $4$.

(c) The graph is the straight line $y=x+2$ with a single point missing — a hole at $(2,4)$, drawn as an open circle. The limit is the height of the hole.

✏️ Example 3 — The value at a does not matter

Let $g(x)=x^2+1$ for all $x$, and let $k(x)=x^2+1$ for $x\ne1$ with $k(1)=5$. Find $\displaystyle\lim_{x\to1}g(x)$, $g(1)$, $\displaystyle\lim_{x\to1}k(x)$ and $k(1)$.

💡 Show solution▶

g For $x$ near $1$, $x^2$ is near $1$, so $g(x)$ is near $2$: $\displaystyle\lim_{x\to1}g(x)=2$, and $g(1)=2$ as well.

k For $x\ne1$, $k(x)=g(x)$, and the limit only looks at $x\ne1$. So $\displaystyle\lim_{x\to1}k(x)=2$ too — but $k(1)=5$.

Lesson Changing (or deleting) the value of a function at the single point $a$ never changes its limit at $a$. The graph of $k$ is the parabola with the point $(1,2)$ lifted to $(1,5)$: a hole at height $2$ and a stray dot at height $5$.

Left-hand and Right-hand Limits

Some functions behave differently on the two sides of a point. The greatest integer function $[x]$ (Section 2.3) is a good example: just below $x=2$, for instance at $x=1.99$, $[x]=1$; just above, at $x=2.01$, $[x]=2$. The graph jumps. To describe such behaviour we look at each side separately.

📘 Definition 12.2 — Left-hand and right-hand limits

The left-hand limit of $f$ at $a$ is the number $l_1$ that $f(x)$ approaches as $x$ tends to $a$ through values less than $a$. We write $$\lim_{x\to a^-}f(x)=l_1\qquad(\text{read "as }x\text{ tends to }a\text{ from the left"}).$$ The right-hand limit $\displaystyle\lim_{x\to a^+}f(x)=l_2$ is the number that $f(x)$ approaches as $x$ tends to $a$ through values greater than $a$. Together they are called the one-sided limits of $f$ at $a$. We often shorten them to LHL and RHL.

📐 Theorem 12.1 — When does a limit exist?

Let $f$ be defined on an open interval containing $a$, except possibly at $a$. Then $\displaystyle\lim_{x\to a}f(x)$ exists and equals $l$ if and only if both one-sided limits exist and are equal to $l$: $$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=l.$$ If the one-sided limits are different, or if either fails to exist, then $\displaystyle\lim_{x\to a}f(x)$ does not exist.

Why. "$x$ close to $a$ on either side" is the same as "$x$ close to $a$ on the left, and $x$ close to $a$ on the right". If $f(x)$ is close to $l$ in both cases it is close to $l$ for all $x$ near $a$; and if $f(x)$ is close to $l$ for all $x$ near $a$ it is so on each side. If the two sides approach different numbers $l_1\ne l_2$, no single $l$ can be close to both, so there is no limit. $\blacksquare$

🧭 Problem-solving strategy — Limits at a point where the formula changes

For the LHL, use the formula that holds for $x$ just below $a$; for the RHL, the formula for $x$ just above $a$. The formula at $x=a$ itself plays no part.
With $[x]$, $|x|$ or $\tfrac{|x|}{x}$, first rewrite the function without brackets on each side (e.g. $[x]=1$ for $1\le x\lt2$, $|x|=-x$ for $x\lt0$).
Compare: equal one-sided limits give the limit; different ones mean the limit does not exist.
Only then, if asked, compare the limit with the value $f(a)$.

✏️ Example 4 — Jumps and corners

Find the left-hand and right-hand limits and decide whether the limit exists: (a) $f(x)=\dfrac{|x|}{x}$ at $x=0$ (b) $[x]$ at $x=2$ and at $x=2.5$ (c) $|x|$ at $x=0$.

💡 Show solution▶

(a) For $x\lt0$, $|x|=-x$, so $f(x)=\tfrac{-x}{x}=-1$; for $x\gt0$, $f(x)=\tfrac xx=1$. LHL $=-1$, RHL $=1$. They differ, so $\displaystyle\lim_{x\to0}\tfrac{|x|}{x}$ does not exist. ($f(0)$ is not defined either, but that is not the reason.)

(b) at 2 For $1\le x\lt2$, $[x]=1$; for $2\le x\lt3$, $[x]=2$. LHL $=1$, RHL $=2$: no limit at $2$, even though $[2]=2$ is defined.

(b) at 2.5 For all $x$ between $2$ and $3$, $[x]=2$. Both one-sided limits are $2$, so $\displaystyle\lim_{x\to2.5}[x]=2$. The greatest integer function has a limit at every non-integer.

(c) For $x\lt0$, $|x|=-x\to0$; for $x\gt0$, $|x|=x\to0$. LHL $=$ RHL $=0$, so $\displaystyle\lim_{x\to0}|x|=0$. The graph has a sharp corner at the origin, but no jump — a corner does not stop the limit from existing.

✏️ Example 5 — Choosing a constant so that the limit exists

(a) $f(x)=\begin{cases}2x+k, & x\lt1\\ x^2+3, & x\ge1.\end{cases}$ Find $k$ so that $\displaystyle\lim_{x\to1}f(x)$ exists.
(b) $g(x)=\begin{cases}ax+1, & x\lt2\\ 5, & x=2\\ x^2-a, & x\gt2.\end{cases}$ Find $a$ so that $\displaystyle\lim_{x\to2}g(x)$ exists. Is the limit equal to $g(2)$?

💡 Show solution▶

(a) LHL $=\displaystyle\lim_{x\to1^-}(2x+k)=2+k$; RHL $=\displaystyle\lim_{x\to1^+}(x^2+3)=4$. The limit exists when $2+k=4$, i.e. $k=2$, and then the limit is $4$.

(b) Sides LHL $=2a+1$; RHL $=4-a$. Setting them equal: $2a+1=4-a$, so $3a=3$ and $a=1$.

(b) Value With $a=1$ both sides approach $3$, so $\displaystyle\lim_{x\to2}g(x)=3$. But $g(2)=5\ne3$: the limit exists and is not the value of the function.

Check $a=1$: $g(1.999)=2.999$ and $g(2.001)=4.004001-1=3.004001$, both close to $3$ ✓.

Pick a function. The red dot moves in towards $x=a$ from the left ($x=a-h$) and the blue dot from the right ($x=a+h$). Press Play to shrink $h$ — the window zooms in with it — or set $h$ with the slider. Watch whether the two heights close in on one number. An open circle is a hole (no value there); a filled dot is the actual value $f(a)$.

function$h$1

$h$—

$f(a-h)$—

$f(a+h)$—

$f(a)$—

limit—

The Algebra of Limits

Tables and graphs are good for guessing a limit; to find limits we need rules. Two limits are the building blocks. A constant function does not change, so $\displaystyle\lim_{x\to a}c=c$. And when $x$ is close to $a$, $x$ is close to $a$, so $\displaystyle\lim_{x\to a}x=a$. The rules below let us build up from these, in the same way that the algebra of real functions (Section 2.4) builds $f+g$, $fg$ and $\tfrac fg$ from $f$ and $g$.

📐 Theorem 12.2 — Algebra of limits

Let $f$ and $g$ be functions such that both $\displaystyle\lim_{x\to a}f(x)=l$ and $\displaystyle\lim_{x\to a}g(x)=m$ exist, and let $\lambda$ be any real number. Then

Sum and difference: $\displaystyle\lim_{x\to a}[f(x)\pm g(x)]=l\pm m$.
Scalar multiple: $\displaystyle\lim_{x\to a}[\lambda f(x)]=\lambda l$.
Product: $\displaystyle\lim_{x\to a}[f(x)\,g(x)]=l\,m$.
Quotient: if also $m\ne0$, then $\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac lm$.

In words: the limit of a sum, difference, multiple, product or quotient is the sum, difference, multiple, product or quotient of the limits — provided the separate limits exist and, for a quotient, the limit of the denominator is not $0$. The same rules hold for left-hand and for right-hand limits.

Why (an intuitive argument). When $x$ is near $a$, $f(x)=l+\text{(small)}$ and $g(x)=m+\text{(small)}$. Then $f(x)+g(x)=l+m+\text{(small)}$, and $f(x)g(x)=lm+l\cdot\text{(small)}$ $+\,m\cdot\text{(small)}+\text{(small)}\cdot\text{(small)}$, which is $lm$ plus something small. A precise proof needs the $\varepsilon$–$\delta$ definition of a limit and is left to a university course.

Repeating the product rule gives $\displaystyle\lim_{x\to a}x^2=a\cdot a=a^2$, then $\displaystyle\lim_{x\to a}x^3=a^3$, and in general $\displaystyle\lim_{x\to a}x^n=a^n$ for every positive integer $n$. Section 12.2 uses this to find the limits of all polynomials and rational functions at once.

✏️ Example 6 — Using the rules

(a) Suppose $\displaystyle\lim_{x\to a}f(x)=3$ and $\displaystyle\lim_{x\to a}g(x)=-2$. Find $\displaystyle\lim_{x\to a}[2f(x)-5g(x)]$, $\displaystyle\lim_{x\to a}f(x)g(x)$, $\displaystyle\lim_{x\to a}\dfrac{f(x)^2}{g(x)+4}$ and $\displaystyle\lim_{x\to a}\dfrac{f(x)+g(x)}{f(x)-g(x)}$.
(b) Find $\displaystyle\lim_{x\to3}(x^2+2x)$, naming the rule used at each step.

💡 Show solution▶

(a) 1 Difference and scalar multiples: $2\times3-5\times(-2)=6+10=16$.

(a) 2 Product: $3\times(-2)=-6$.

(a) 3 The denominator tends to $-2+4=2\ne0$, so the quotient rule applies: $\dfrac{3^2}{2}=\dfrac92$.

(a) 4 The denominator tends to $3-(-2)=5\ne0$: $\dfrac{3+(-2)}{5}=\dfrac15$.

(b) $\displaystyle\lim_{x\to3}(x^2+2x)=\lim_{x\to3}x\cdot\lim_{x\to3}x+2\lim_{x\to3}x$ (sum, product and scalar-multiple rules) $=3\cdot3+2\cdot3=15$ (using $\displaystyle\lim_{x\to3}x=3$).

Check $x=3.001$: $9.006001+6.002=15.008001$; $x=2.999$: $8.994001+5.998=14.992001$ — both near $15$ ✓.

Each function is given by one formula to the left of $a$ and another to the right. The left formula contains a constant $k$. Drag the $k$ slider until the left-hand limit equals the right-hand limit; Play sends the two dots in towards $a$ so you can watch where each side is heading. Only when the gap is $0$ does $\displaystyle\lim_{x\to a}f(x)$ exist.

function$k$−1.5

LHL—

RHL—

gap—

$\lim_{x\to a}f(x)$—

⚠️ Common mistakes with limits

1. "The limit is just $f(a)$." Often it is, but not always: in Example 3 the limit is $2$ while $k(1)=5$, and in Example 2 $f(2)$ does not even exist. Always think about $x$ near $a$.

2. "$f(a)$ is undefined, so there is no limit." A hole in the graph does not stop the limit from existing. $\dfrac{x^2-4}{x-2}\to4$ as $x\to2$.

3. Checking only one side. A table that uses only $x=1.9,\ 1.99,\ 1.999$ finds the left-hand limit. For the limit you must also approach from the right.

4. Using the formula at $x=a$ for a one-sided limit. For $f(x)=x+2$ when $x\ge1$ and $f(x)=x+1$ when $x\lt1$, the left-hand limit at $1$ comes from $x+1$, giving $2$ — not from the value $f(1)=3$.

5. Splitting a limit that does not exist. Theorem 12.2 needs both separate limits to exist. $\displaystyle\lim_{x\to0}\left(\tfrac{|x|}{x}-\tfrac{|x|}{x}\right)=0$, but you cannot write it as $\displaystyle\lim\tfrac{|x|}{x}-\lim\tfrac{|x|}{x}$, because neither limit exists.

6. Dividing when the denominator tends to $0$. The quotient rule needs $m\ne0$. Limits such as $\dfrac{x^2-4}{x-2}$ at $2$ (numerator and denominator both $\to0$) need algebra first — the subject of Section 12.2.

Exercises

Limits from tables and the meaning of a limit.

1 For $f(x)=3x-1$, find $f(x)$ at $x=1.9,\ 1.99,\ 2.01,\ 2.1$ and estimate $\displaystyle\lim_{x\to2}f(x)$.
2 Use a table of values on both sides to estimate $\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}$. Is the function defined at $x=3$?
3 Find $\dfrac{x^3-1}{x-1}$ at $x=0.9,\ 0.99,\ 1.01,\ 1.1$ (to four decimal places) and estimate its limit as $x\to1$.
4 Simplify $\dfrac{(1+x)^2-1}{x}$ for $x\ne0$ and hence find its limit as $x\to0$.
5 Estimate $\displaystyle\lim_{x\to4}\dfrac{\sqrt x-2}{x-4}$ from the values at $x=3.9,\ 3.99,\ 4.01,\ 4.1$ (to four decimal places).
6 Let $f(x)=x+1$ for $x\ne2$ and $f(2)=7$. Find $\displaystyle\lim_{x\to2}f(x)$ and $f(2)$, and sketch the graph near $x=2$.
7 True or false? "If $f(a)$ is not defined, then $\displaystyle\lim_{x\to a}f(x)$ does not exist." Give a reason.
8 True or false? "If $\displaystyle\lim_{x\to a}f(x)=l$, then $f(a)=l$." Give a reason.
9 Estimate $\displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{x+1}-1}$ from the values at $x=\pm0.1$ and $x=\pm0.01$.

Check: a table of values on both sides of $a$

Part:

💡 Show solutions▶

1 $4.7,\ 4.97,\ 5.03,\ 5.3$; the limit is $5$.

2 $x=2.9,\ 2.99,\ 3.01,\ 3.1$ give $5.9,\ 5.99,\ 6.01,\ 6.1$ (the expression equals $x+3$ for $x\ne3$): limit $6$. It is not defined at $x=3$ ($\tfrac00$).

3 $2.71,\ 2.9701,\ 3.0301,\ 3.31$; the limit is $3$. (For $x\ne1$ the expression is $x^2+x+1$.)

4 $\dfrac{1+2x+x^2-1}{x}=\dfrac{2x+x^2}{x}=2+x$ for $x\ne0$, so the limit is $2$.

5 $0.2516,\ 0.2502,\ 0.2498,\ 0.2485$; the limit is $0.25=\tfrac14$.

6 Near $2$ (not at $2$), $f(x)=x+1\to3$: the limit is $3$, while $f(2)=7$. The graph is the line $y=x+1$ with a hole at $(2,3)$ and a filled dot at $(2,7)$.

7 False. $\dfrac{x^2-4}{x-2}$ is not defined at $2$ but its limit there is $4$ (Example 2).

8 False. The limit ignores $x=a$: in part 6 the limit is $3$ but $f(2)=7$; and $f(a)$ may not exist at all.

9 $x=0.1$: $2.0488$; $x=-0.1$: $1.9487$; $x=0.01$: $2.0050$; $x=-0.01$: $1.9950$. The limit is $2$. (In fact the expression equals $\sqrt{x+1}+1$ for $x\ne0$, $x\ge-1$.)

Left-hand and right-hand limits.

1 Find $\displaystyle\lim_{x\to0^-}\dfrac{|x|}{x}$ and $\displaystyle\lim_{x\to0^+}\dfrac{|x|}{x}$.
2 Find $\displaystyle\lim_{x\to3^-}[x]$ and $\displaystyle\lim_{x\to3^+}[x]$. Does $\displaystyle\lim_{x\to3}[x]$ exist?
3 Find $\displaystyle\lim_{x\to1.5}[x]$.
4 $f(x)=x^2$ for $x\le1$ and $f(x)=2x$ for $x\gt1$. Find the one-sided limits at $x=1$. Does the limit exist?
5 $f(x)=3x+1$ for $x\lt2$ and $f(x)=x^2+3$ for $x\ge2$. Find $\displaystyle\lim_{x\to2}f(x)$ if it exists.
6 $f(x)=kx^2$ for $x\lt1$ and $f(x)=3x-1$ for $x\ge1$. Find $k$ so that $\displaystyle\lim_{x\to1}f(x)$ exists.
7 Find the one-sided limits of $\dfrac{|x-2|}{x-2}$ at $x=2$.
8 Find the one-sided limits of $f(x)=x-[x]$ at $x=1$. Does $\displaystyle\lim_{x\to1}f(x)$ exist?
9 $f(x)=a+2bx$ for $x\lt1$, $f(1)=5$, and $f(x)=3b-ax$ for $x\gt1$. Find $a$ and $b$ if $\displaystyle\lim_{x\to1}f(x)=f(1)$.
10 Using the signum function ($\operatorname{sgn}x=-1,0,1$ for $x\lt0$, $x=0$, $x\gt0$), decide whether $\displaystyle\lim_{x\to0}\operatorname{sgn}x$ and $\displaystyle\lim_{x\to0}(\operatorname{sgn}x)^2$ exist, and compare the second with $(\operatorname{sgn}0)^2$.

💡 Show solutions▶

1 $-1$ and $1$ (for $x\lt0$ the fraction is $\tfrac{-x}{x}=-1$; for $x\gt0$ it is $1$).

2 $2$ and $3$. They differ, so the limit does not exist.

3 $[x]=1$ for all $x$ in $[1,2)$, an interval containing $1.5$, so the limit is $1$.

4 LHL $=1^2=1$, RHL $=2\times1=2$. Different: the limit does not exist.

5 LHL $=3\times2+1=7$, RHL $=4+3=7$. Equal, so the limit is $7$ (and it equals $f(2)=7$).

6 LHL $=k$, RHL $=3-1=2$, so $k=2$.

7 For $x\lt2$, $|x-2|=-(x-2)$, giving $-1$; for $x\gt2$ it gives $1$. LHL $=-1$, RHL $=1$.

8 For $0\le x\lt1$, $[x]=0$ and $f(x)=x\to1$; for $1\le x\lt2$, $[x]=1$ and $f(x)=x-1\to0$. LHL $=1$, RHL $=0$: no limit.

9 Both one-sided limits must equal $5$: $a+2b=5$ and $3b-a=5$. Adding, $5b=10$, so $b=2$ and $a=1$. Check: $1+4=5$ and $6-1=5$ ✓.

10 $\displaystyle\lim_{x\to0}\operatorname{sgn}x$ does not exist (LHL $-1$, RHL $1$). But $(\operatorname{sgn}x)^2=1$ for every $x\ne0$, so $\displaystyle\lim_{x\to0}(\operatorname{sgn}x)^2=1$, while $(\operatorname{sgn}0)^2=0$. (A product of two functions can have a limit even when the factors do not — Theorem 12.2 works only in the other direction.)

III

The algebra of limits. In parts 1–6, $\displaystyle\lim_{x\to a}f(x)=4$, $\displaystyle\lim_{x\to a}g(x)=-1$ and $\displaystyle\lim_{x\to a}p(x)=0$.

1 Find $\displaystyle\lim_{x\to a}[3f(x)+2g(x)]$.
2 Find $\displaystyle\lim_{x\to a}f(x)g(x)$.
3 Find $\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}$.
4 Find $\displaystyle\lim_{x\to a}[f(x)^2-g(x)^2]$.
5 Find $\displaystyle\lim_{x\to a}\dfrac{f(x)+p(x)}{g(x)-2}$.
6 Explain why Theorem 12.2 cannot be used to find $\displaystyle\lim_{x\to a}\dfrac{f(x)}{p(x)}$.
7 Use the rules, step by step, to find $\displaystyle\lim_{x\to-2}(x^3+4x)$.
8 Find $\displaystyle\lim_{x\to1}(2x+3)(x^2-5)$.

💡 Show solutions▶

1 $3\times4+2\times(-1)=10$.

2 $4\times(-1)=-4$.

3 The denominator's limit is $-1\ne0$: $\tfrac{4}{-1}=-4$.

4 $16-1=15$.

5 The denominator tends to $-3\ne0$: $\dfrac{4+0}{-3}=-\dfrac43$.

6 The quotient rule needs the denominator's limit to be non-zero, but $p(x)\to0$. (In fact the fraction cannot have a finite limit here: its numerator tends to $4$ while its denominator tends to $0$, so its size grows without bound.)

7 $\displaystyle\lim x^3=(-2)^3=-8$ (product rule twice) and $\displaystyle\lim4x=4\times(-2)=-8$ (scalar multiple), so the sum rule gives $-16$.

8 $\displaystyle\lim(2x+3)=5$ and $\displaystyle\lim(x^2-5)=-4$, so by the product rule the limit is $-20$.

Rates of change and applications.

1 A stone dropped from a cliff falls $s=4.9t^2$ metres in $t$ seconds. Find its average speed over $[3,3.1]$ and over $[3,3.01]$, and its instantaneous speed at $t=3$.
2 A cyclist is $s=t^2+3t$ metres from the start after $t$ seconds. Show that the average speed over $[2,2+h]$ is $7+h$ and find the speed at $t=2$.
3 A square metal plate of side $x\ \text{cm}$ is heated and expands. Find the average rate of change of its area $A=x^2$ as $x$ changes from $5$ to $5+h$, and the limit of this rate as $h\to0$, with units.
4 A tank holds $V=1000-20t+0.1t^2$ litres of water $t$ minutes after a tap is opened ($0\le t\le100$). Find the average rate of change of $V$ over $[10,10+h]$ and its limit as $h\to0$. Interpret the sign.
5 A car park charges $C(t)=30+20[t]$ rupees for parking $t$ hours ($0\le t\lt5$). Find $\displaystyle\lim_{t\to2^-}C(t)$ and $\displaystyle\lim_{t\to2^+}C(t)$, and say what the difference means to a driver.

💡 Show solutions▶

1 Over $[3,3+h]$ the average speed is $\dfrac{4.9(6h+h^2)}{h}=29.4+4.9h$: $29.89\ \text{m/s}$ for $h=0.1$, $29.449\ \text{m/s}$ for $h=0.01$. As $h\to0$ it tends to $29.4\ \text{m/s}$.

2 $s(2)=10$, $s(2+h)=4+4h+h^2+6+3h$ $=10+7h+h^2$, so the average speed is $\dfrac{7h+h^2}{h}=7+h$ ($h\ne0$). The speed at $t=2$ is $7\ \text{m/s}$.

3 $\dfrac{(5+h)^2-25}{h}=\dfrac{10h+h^2}{h}=10+h\ \text{cm}^2$ per cm; the limit is $10\ \text{cm}^2$ per cm of side.

4 $V(10)=810$ and $V(10+h)=810-18h+0.1h^2$, so the average rate is $-18+0.1h$ litres per minute, with limit $-18\ \text{L/min}$. The minus sign means the volume is decreasing: at $t=10$ the tank is emptying at $18$ litres per minute.

5 Just under $2$ hours $[t]=1$: $C\to50$. Just over $2$ hours $[t]=2$: $C\to70$. The one-sided limits are ₹50 and ₹70, so $\displaystyle\lim_{t\to2}C(t)$ does not exist: one minute past the two-hour mark costs an extra ₹20.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for one-sided limits of piecewise functions, $|x|$ and $[x]$ (1–3 marks), and for constants that make a limit exist or make it equal the value of the function. JEE combines the algebra of limits with greatest-integer and modulus functions, and likes multiple-correct questions on whether a limit exists and whether it equals $f(a)$.

E1. Board · 1 mark $\displaystyle\lim_{x\to0}\dfrac{x}{|x|}$ is (A) $1$ (B) $-1$ (C) $0$ (D) does not exist

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Idea For $x\gt0$ the fraction is $1$; for $x\lt0$ it is $\tfrac{x}{-x}=-1$. LHL $=-1\ne1=$ RHL.

Trap (A) checks only the right side (for example with a table using $x=0.1,0.01$); (C) treats $\tfrac00$ as $0$.

Answer (D).

E2. Board · 2 marks $f(x)=\begin{cases}x^2+1, & x\le2\\ 3x-1, & x\gt2.\end{cases}$ Find $\displaystyle\lim_{x\to2}f(x)$, if it exists, and compare it with $f(2)$.

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Sides LHL $=\displaystyle\lim_{x\to2^-}(x^2+1)=5$; RHL $=\displaystyle\lim_{x\to2^+}(3x-1)=5$. (1 mark)

Answer Equal, so $\displaystyle\lim_{x\to2}f(x)=5$; and $f(2)=2^2+1=5$, the same. (1 mark)

E3. Board · 3 marks $f(x)=\begin{cases}x+a, & x\lt0\\ x^2+1, & 0\le x\lt1\\ bx-2, & x\ge1.\end{cases}$ Find $a$ and $b$ so that $\displaystyle\lim_{x\to0}f(x)$ and $\displaystyle\lim_{x\to1}f(x)$ both exist.

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At 0 LHL $=a$, RHL $=0+1=1$, so $a=1$. (1 mark)

At 1 LHL $=1+1=2$, RHL $=b-2$, so $b-2=2$ and $b=4$. (1 mark)

Answer $a=1$, $b=4$; the limits are $1$ at $x=0$ and $2$ at $x=1$. (1 mark)

E4. Board · Case study A stone is dropped from a bridge. Taking $g\approx10\ \text{m/s}^2$, it falls $s=5t^2$ metres in the first $t$ seconds. (i) Find its average speed during the first $2$ seconds. (1 mark) (ii) Show that its average speed over $[1,1+h]$ is $10+5h$ m/s. (1 mark) (iii) Find its speed at $t=1\ \text{s}$, and the time at which its speed is $25\ \text{m/s}$. (2 marks)

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(i) $\dfrac{s(2)-s(0)}{2}=\dfrac{20}{2}=10\ \text{m/s}$. (1 mark)

(ii) $\dfrac{5(1+h)^2-5}{h}=\dfrac{10h+5h^2}{h}=10+5h$ for $h\ne0$. (1 mark)

(iii) at t = 1 As $h\to0$, $10+5h\to10$: the speed at $t=1$ is $10\ \text{m/s}$. (1 mark)

(iii) 25 m/s In the same way, over $[a,a+h]$ the average is $\dfrac{5(2ah+h^2)}{h}=10a+5h\to10a$. So the speed at time $a$ is $10a$, and $10a=25$ gives $a=2.5\ \text{s}$. (1 mark)

E5. JEE Main If $\displaystyle\lim_{x\to a}f(x)=2$ and $\displaystyle\lim_{x\to a}g(x)=-3$, then $\displaystyle\lim_{x\to a}\dfrac{f(x)^2+g(x)}{f(x)-2g(x)}$ equals (A) $\tfrac18$ (B) $-\tfrac14$ (C) $\tfrac{13}{8}$ (D) $\tfrac12$

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Method Numerator $\to4+(-3)=1$; denominator $\to2-2(-3)=8\ne0$. By the quotient rule the limit is $\tfrac18$.

Trap (B) comes from the sign slip $2-6=-4$ in the denominator; (C) squares $g$ as well, $4+9=13$.

Answer (A).

E6. JEE Advanced One or more options correct. Let $f(x)=[x]+[-x]$, where $[\ ]$ is the greatest integer function. Then (A) $\displaystyle\lim_{x\to2}f(x)$ exists and equals $-1$ (B) $f(2)=0$ (C) $\displaystyle\lim_{x\to2}f(x)=f(2)$ (D) $\displaystyle\lim_{x\to2.5}f(x)=-1$

💡 Show solution▶

Left of 2 For $1\lt x\lt2$: $[x]=1$ and $-2\lt-x\lt-1$, so $[-x]=-2$. Sum $-1$.

Right of 2 For $2\lt x\lt3$: $[x]=2$ and $-3\lt-x\lt-2$, so $[-x]=-3$. Sum $-1$. Both one-sided limits are $-1$, so (A) is true.

At 2 and 2.5 $f(2)=2+(-2)=0$, so (B) is true and (C) is false. Near $2.5$ (inside $(2,3)$) $f(x)=-1$, so (D) is true.

⏱ Fast route $[x]+[-x]=0$ when $x$ is an integer and $-1$ otherwise; the limit at any point is $-1$.

Answer (A), (B) and (D).

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. How fast is a ball rolling down a ramp at one particular instant — and how short must a timing interval be before the average speed is a good estimate of that instantaneous speed?

Why it matters. Every speed measurement is secretly an average over a short interval: a cricket speed gun, a traffic radar, a phone's GPS speed and the timing gates of an athletics track all divide a small distance by a small time. Engineers have to choose the interval: too long and the speed changes during it; too short and the measuring errors swamp the tiny distance. The limit is the ideal they are trying to approach.

What to do.

Make a gentle ramp (a long plank or a grooved ruler raised at one end) and stick a tape measure along it. Release a ball from rest at the top and film it with a phone, ideally in slow motion (many phones film at $120$ or $240$ frames per second).
Step through the video frame by frame and record the position $s$ of the ball at several times $t$ (the time between frames is $\tfrac1{120}\ \text{s}$ or $\tfrac1{240}\ \text{s}$).
Choose an instant $t_0$ near the middle of the run. Compute average speeds $\dfrac{s(t_0+h)-s(t_0)}{h}$ for $h$ equal to $0.5$, $0.2$, $0.1$, $0.05$ and one frame, on both sides of $t_0$ ($h\gt0$ and $h\lt0$).
Tabulate the averages and see whether they settle. Estimate the instantaneous speed at $t_0$.
Fit a model $s=ct^2$ to your data in a spreadsheet; for this model the speed at $t_0$ is $2ct_0$ (you will see why in Section 12.3). Compare.

Tools: A phone camera (slow motion if available), a plank or ruler, a tape measure, a spreadsheet

Hand in: Your table of positions, the table of average speeds for shrinking $h$ on both sides, your estimate of the speed at $t_0$, and a paragraph on why the smallest $h$ did not necessarily give the best estimate.

Going further. Repeat with the ramp twice as steep. How does $c$ change? Look up how an athletics photo-finish camera or a traffic radar works and describe which "$h$" it uses.

Instructor note: for a ball rolling on a ramp of angle $\theta$, $s\approx\tfrac12\cdot\tfrac57 g\sin\theta\,t^2$ (rolling, not sliding), so with $\theta=5^\circ$, $c\approx0.30\ \text{m/s}^2$. With one-frame intervals the reading error in $s$ (a few millimetres) is divided by a very small $h$, so the one-frame averages scatter by tens of centimetres per second — a good discussion point about why the mathematical limit and a physical measurement differ.

🧪 Self-check quiz

0/5 correct

Q1. $\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}$ is

Q2. $\displaystyle\lim_{x\to4^-}[x]$ is

Q3. Which statement is always true?

Q4. If $\displaystyle\lim_{x\to a}f(x)=5$ and $\displaystyle\lim_{x\to a}g(x)=2$, then $\displaystyle\lim_{x\to a}\dfrac{f(x)-g(x)}{f(x)+g(x)}$ is

Q5. $f(x)=x+k$ for $x\lt1$ and $f(x)=3x$ for $x\ge1$. The limit of $f$ at $1$ exists when $k$ equals

← Chapter 12 — Limits and Derivatives 12.2 Limits of Polynomial, Rational and Trigonometric Functions →