Chapter 1 · Limits and Continuity

1.3 Continuity

Three conditions decide whether a graph can be traced through a point without lifting the pencil — and they are all built from limits.

Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are called continuous. Other functions have points at which a break occurs, but satisfy the property over intervals contained in their domains: they are continuous on those intervals and are said to have a discontinuity at a point where a break occurs.

Continuity at a Point

Before the formal definition, look at functions that fail our intuitive notion of continuity at a point. Each failure adds one condition to the list.

Figure 1.4 — not continuous at $a$ because $f(a)$ is undefined.
Figure 1.5 — not continuous at $a$ because $\displaystyle\lim_{x\to a}f(x)$ does not exist.
Figure 1.6 — not continuous at $a$ because $\displaystyle\lim_{x\to a}f(x)\neq f(a)$.
📘 Definition — Continuous at a point
A function is continuous at a point $a$ if and only if all three of the following are satisfied:
  1. $f(a)$ is defined;
  2. $\displaystyle\lim_{x\to a}f(x)$ exists;
  3. $\displaystyle\lim_{x\to a}f(x)=f(a)$.
A function is discontinuous at $a$ if it fails to be continuous at $a$.
Interactive 1 — The Three-Condition Checker
from the left—
from the right—
$f(a)$—
?1. $f(a)$ is defined
?2. the limit exists
?3. limit $=f(a)$
✏️ Example 1 — Determining continuity at a point
Using the definition, determine whether $f(x)=\left\{\begin{array}{cl} \dfrac{\sin x}{x} & \quad x\neq0 \\[6pt] 1 & \quad x=0 \end{array}\right.$ is continuous at $x=0$. Justify the conclusion.
Condition 1 First observe that $f(0)=1$, so the value is defined.
Condition 2 Next, $$\lim_{x\to0}f(x)=\lim_{x\to0}\frac{\sin x}{x}=1,$$ a limit established numerically in Exercise 3 of Section 1.1. So the limit exists.
Condition 3 Compare the two: $\displaystyle\lim_{x\to0}f(x)=1=f(0)$.
Conclusion All three conditions hold, so $f$ is continuous at $x=0$. Notice what the piecewise value did: the formula $\tfrac{\sin x}{x}$ alone has a hole at $0$, and defining $f(0)=1$ fills it at exactly the right height.
✏️ Example 2 — Determining continuity at a point
Using the definition, determine whether $f(x)=\left\{\begin{array}{cl} -x^2+4 & \quad x\le3 \\[2pt] 4x-8 & \quad x>3 \end{array}\right.$ is continuous at $x=3$. Justify the conclusion.
Condition 1 Begin by calculating $f(3)$. Since $3\le3$ we use the first branch: $$f(3)=-(3)^2+4=-5,$$ so $f(3)$ is defined.
Condition 2 Next compute the two one-sided limits: $$\lim_{x\to3^-}f(x)=-(3)^2+4=-5,\qquad \lim_{x\to3^+}f(x)=4(3)-8=4.$$ They differ, so $\displaystyle\lim_{x\to3}f(x)$ does not exist.
Conclusion Condition 2 fails, so $f$ is not continuous at $3$. This is a jump discontinuity: no way to redefine $f(3)$ can repair it, because the two sides do not agree with each other.
📐 Theorem 1.5 — Continuity of polynomials and rational functions
Polynomials and rational functions are continuous at every point in their domains.
📐 Theorem 1.6 — Continuity of trigonometric functions
Trigonometric functions are continuous over their entire domains.
✏️ Example 3 — Continuity of a rational function
For what values of $x$ is $f(x)=\dfrac{x+1}{x-5}$ continuous?
Step 1 By Theorem 1.5 a rational function is continuous everywhere in its domain, so the only possible trouble is where the denominator vanishes: $x-5=0$, that is $x=5$.
Step 2 The function is therefore continuous for every value of $x$ except $x=5$ — in interval language, on $(-\infty,5)\cup(5,\infty)$.
Note At $x=5$ the function is not merely discontinuous, it is undefined: condition 1 fails. There is a vertical asymptote there, so no redefinition can rescue it.

Continuity over an Interval

Now we extend the idea from a point to an interval. Keep the intuition in mind: a function is continuous over an interval if you can trace it between any two points of the interval without lifting the pencil.

📘 Definition — Continuity from the right and from the left
A function $f(x)$ is continuous from the right at $a$ if $\displaystyle\lim_{x\to a^+}f(x)=f(a)$.
A function $f(x)$ is continuous from the left at $a$ if $\displaystyle\lim_{x\to a^-}f(x)=f(a)$.

A function is continuous over an open interval if it is continuous at every point of the interval. It is continuous over a closed interval $[a,b]$ if it is continuous at every point of $(a,b)$, continuous from the right at $a$, and continuous from the left at $b$. Intervals of the form $(a,b]$ and $[a,b)$ are handled in the same fashion.

Interactive 2 — The Pencil Test

Watch the pencil trace the curve from left to right. Every time it has to lift, you have found a break — and the intervals between the lifts are the intervals of continuity.

pencil position $x$—
status—
lifts so far0
✏️ Example 4 — Continuity over an interval
State the interval(s) over which the function $f(x)=\dfrac{x-1}{x^2+2x}$ is continuous.
Step 1 Since $f$ is a rational function, Theorem 1.5 says it is continuous at every point of its domain. So the work is finding the domain.
Step 2 The denominator factors as $x^2+2x=x(x+2)$, which vanishes at $x=0$ and $x=-2$. The domain is $$(-\infty,-2)\cup(-2,0)\cup(0,\infty).$$
Step 3 Therefore $f$ is continuous over each of the intervals $(-\infty,-2)$, $(-2,0)$ and $(0,\infty)$.
Careful Do not write the answer as one interval with points removed and a union sign misused: continuity is claimed on each interval separately, because the function is not defined at the two excluded points at all.
✏️ Example 5 — Continuity over an interval
State the interval(s) over which the function $f(x)=\sqrt{4-x^2}$ is continuous.
Step 1 From the limit laws (root law), $\displaystyle\lim_{x\to a}\sqrt{4-x^2}=\sqrt{4-a^2}$ for all $a$ in $(-2,2)$, so $f$ is continuous at every interior point.
Step 2 At the ends we can only approach from inside the domain: $$\lim_{x\to-2^+}\sqrt{4-x^2}=0=f(-2),\qquad \lim_{x\to2^-}\sqrt{4-x^2}=0=f(2),$$ so $f$ is continuous from the right at $-2$ and continuous from the left at $2$.
Conclusion Therefore $f$ is continuous over the closed interval $[-2,2]$.
Why one-sided is enough Outside $[-2,2]$ the radicand is negative and $f$ is not defined, so a two-sided limit at an endpoint is not even meaningful. This is exactly why the definition allows one-sided continuity at endpoints.

Exercises

The Section 1.3 exercises come first (I–IV), followed by additional practice (V–VIII).

I
Using the definition, determine whether $f(x)=\left\{\begin{array}{cl} 2x+1 & \quad x<1 \\[2pt] 2 & \quad x=1 \\[2pt] -x+4 & \quad x>1 \end{array}\right.$ is continuous at $x=1$. Justify the conclusion.
Check the three conditions
from the left—
from the right—
$f(1)$—
?1. $f(1)$ defined
?2. limit exists
?3. limit $= f(1)$
Condition 1 The middle line of the definition gives $f(1)=2$ directly, so $f(1)$ is defined. ✅
Condition 2 One-sided limits use the outer branches: $$\lim_{x\to1^-}f(x)=2(1)+1=3,\qquad \lim_{x\to1^+}f(x)=-(1)+4=3.$$ They agree, so $\displaystyle\lim_{x\to1}f(x)=3$ exists. ✅
Condition 3 Compare: the limit is $3$ but $f(1)=2$, so $\displaystyle\lim_{x\to1}f(x)\neq f(1)$. ❌
Conclusion Condition 3 fails, so $f$ is not continuous at $x=1$. The discontinuity is removable: redefining $f(1)=3$ would make it continuous, because the two sides already agree with each other.
II
State the interval(s) over which the function $f(x)=\sqrt{x+3}$ is continuous.
Pencil test on $\sqrt{x+3}$
pencil position $x$—
status—
lifts so far0
Step 1 The domain needs $x+3\ge0$, that is $x\ge-3$. Outside that the function does not exist, so continuity cannot even be discussed there.
Step 2 On $(-3,\infty)$ the root law gives $\displaystyle\lim_{x\to a}\sqrt{x+3}=\sqrt{a+3}=f(a)$, so $f$ is continuous at every interior point.
Step 3 At the left endpoint, $\displaystyle\lim_{x\to-3^+}\sqrt{x+3}=0=f(-3)$, so $f$ is continuous from the right at $-3$.
Conclusion $f$ is continuous over $\left[-3,\infty\right)$.

III. Continuity at a given point

III
Determine whether the function is continuous at the given point.
  • 1 $f(x)=\dfrac{2x^2-5x+3}{x-1}$ at $x=1$
  • 2 $f(x)=\left\{\begin{array}{cl} x\sin x & \quad x\le\pi \\[2pt] x\tan x & \quad x>\pi \end{array}\right.$ at $x=\pi$
  • 3 $h(x)=\left\{\begin{array}{cl} x^2-e^x & \quad x<0 \\[2pt] x-1 & \quad x\ge0 \end{array}\right.$ at $x=0$
Check the three conditions
from the left—
from the right—
$f$ at the point—
?1. value defined
?2. limit exists
?3. limit = value
1 At $x=1$ the denominator is $0$, so $f(1)$ is undefined and condition 1 already fails: $f$ is not continuous at $1$. The limit itself does exist — factoring gives $\dfrac{(2x-3)(x-1)}{x-1}=2x-3\to-1$ — so the discontinuity is removable, and defining $f(1)=-1$ would repair it.
2 Condition 1: $f(\pi)=\pi\sin\pi=0$. Condition 2: from the left, $x\sin x\to\pi\sin\pi=0$; from the right, $x\tan x\to\pi\tan\pi=0$. Both sides give $0$, so the limit exists and equals $0$. Condition 3: $0=f(\pi)$. All three hold, so $f$ is continuous at $\pi$.
3 Condition 1: $h(0)=0-1=-1$. Condition 2: from the left, $x^2-e^x\to0-1=-1$; from the right, $x-1\to-1$. The limit is $-1$. Condition 3: $-1=h(0)$. So $h$ is continuous at $0$.
Pattern For a piecewise function, evaluate each branch at the break point. If the two branch values agree with one another and with the defined value, the function is continuous there.
III
Continued.
  • 4 $g(u)=\left\{\begin{array}{cl} \dfrac{6u^2+u-2}{2u-1} & \quad u\neq\tfrac12 \\[6pt] \dfrac72 & \quad u=\tfrac12 \end{array}\right.$ at $u=\tfrac12$
  • 5 $f(y)=\dfrac{\sin(\pi y)}{\tan(\pi y)}$ at $y=1$
Check the three conditions
from the left—
from the right—
value at the point—
?1. value defined
?2. limit exists
?3. limit = value
4 Condition 1: the second line defines $g\!\left(\tfrac12\right)=\tfrac72$. ✅ Condition 2: factor the numerator, $6u^2+u-2=(2u-1)(3u+2)$, so for $u\neq\tfrac12$, $$g(u)=\frac{(2u-1)(3u+2)}{2u-1}=3u+2\ \Longrightarrow\ \lim_{u\to\frac12}g(u)=3\left(\tfrac12\right)+2=\tfrac72.$$ ✅ Condition 3: $\tfrac72=g\!\left(\tfrac12\right)$. ✅ So $g$ is continuous at $u=\tfrac12$ — the patch was chosen at exactly the right height.
5 Condition 1: at $y=1$, $\tan(\pi)=0$, so $f(1)$ is a division by zero and is undefined. ❌ The function is not continuous at $y=1$.
5, continued The limit nevertheless exists: for $y$ near $1$, $$\frac{\sin(\pi y)}{\tan(\pi y)}=\frac{\sin(\pi y)\cos(\pi y)}{\sin(\pi y)}=\cos(\pi y)\ \longrightarrow\ \cos\pi=-1.$$ So the discontinuity is removable, and setting $f(1)=-1$ would repair it.
Contrast Parts 4 and 5 are the same situation with different endings: in part 4 the hole is patched at the correct height, in part 5 nobody patched it at all.

IV. Choosing $k$ to force continuity

IV
Find the value(s) of $k$ that make each function continuous over the given interval.
  • 1 $f(x)=\left\{\begin{array}{cl} \dfrac{x^2+3x+2}{x+2} & \quad x\neq-2 \\[6pt] k & \quad x=-2 \end{array}\right.$
  • 2 $g(x)=\left\{\begin{array}{cl} \sqrt{kx} & \quad 0\le x\le3 \\[2pt] x+1 & \quad 3\le x\le10 \end{array}\right.$
Slide $k$ until the pieces join
current $k$—
left value—
right value—
gap—
1 For $x\neq-2$, factor: $\dfrac{(x+1)(x+2)}{x+2}=x+1$, so $$\lim_{x\to-2}f(x)=-2+1=-1.$$ Continuity needs the value to equal the limit, so $k=-1$.
2 The two rules must agree where the intervals meet, at $x=3$: $$\sqrt{3k}=3+1=4\ \Longrightarrow\ 3k=16\ \Longrightarrow\ k=\frac{16}{3}.$$ Each piece is continuous on its own interval, so matching the values at the junction is all that is required.
Method In every problem of this type: compute the limit from each side in terms of $k$, set them equal (and equal to the defined value), then solve for $k$.
IV
Continued.
  • 3 $g(x)=\left\{\begin{array}{cl} e^{kx} & \quad 0\le x\le4 \\[2pt] x+3 & \quad 4\le x\le8 \end{array}\right.$
  • 4 $h(x)=\left\{\begin{array}{cl} 3x+2 & \quad x>4 \\[2pt] 2x-k & \quad x\le4 \end{array}\right.$
Slide $k$ until the pieces join
current $k$—
left value—
right value—
gap—
3 Match the two rules at $x=4$: $$e^{4k}=4+3=7\ \Longrightarrow\ 4k=\ln 7\ \Longrightarrow\ k=\frac{\ln 7}{4}\approx0.4865.$$
4 Here the value at $4$ comes from the second line, $h(4)=8-k$, and the right-hand limit comes from the first: $$\lim_{x\to4^+}h(x)=3(4)+2=14.$$ Continuity needs $8-k=14$, so $k=-6$.
Check With $k=-6$ the lower branch is $2x+6$, which at $x=4$ gives $14$ — the same height the upper branch approaches. ✅

V. Classifying discontinuities

V
Each function is discontinuous at the given point. Say which of the three conditions fails and classify the discontinuity as removable, jump or infinite.
  • 1 $f(x)=\dfrac{x^2-9}{x-3}$ at $x=3$
  • 2 $f(x)=\dfrac{|x-2|}{x-2}$ at $x=2$
  • 3 $f(x)=\dfrac{1}{(x+1)^2}$ at $x=-1$
  • 4 $f(x)=\left\{\begin{array}{cl} x^2+1 & \quad x<1 \\[2pt] 3 & \quad x=1 \\[2pt] 2x & \quad x>1 \end{array}\right.$ at $x=1$
  • 5 Applied. A car park charges $2$ OMR for every hour or part of an hour, so the fee after $t$ hours is $P(t)=2\lceil t\rceil$ for $0<t\le4$ (for example $P(1.2)=4$ and $P(2)=4$). Classify the discontinuity at $t=2$ and explain what it means for a driver.
Which condition breaks?
from the left—
from the right—
value at the point—
?1. value defined
?2. limit exists
?3. limit = value
1 $f(3)$ is $\tfrac00$, so condition 1 fails. For $x\neq3$, $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3\ \longrightarrow\ 6.$$ The limit exists, so the discontinuity is removable (define $f(3)=6$).
2 $f(2)$ is $\tfrac00$, undefined. For $x<2$, $|x-2|=-(x-2)$ so $f(x)=-1$; for $x>2$, $f(x)=1$. Hence $$\lim_{x\to2^-}f(x)=-1,\qquad \lim_{x\to2^+}f(x)=1.$$ The one-sided limits are finite but different: a jump discontinuity of size $2$. Conditions 1 and 2 both fail.
3 $f(-1)$ is undefined, and as $x\to-1$ from either side $(x+1)^2\to0^+$, so $$\lim_{x\to-1}\frac{1}{(x+1)^2}=+\infty.$$ The limit does not exist as a real number: an infinite discontinuity, with vertical asymptote $x=-1$.
4 $f(1)=3$ is defined. From the left $x^2+1\to2$, from the right $2x\to2$, so $\displaystyle\lim_{x\to1}f(x)=2$ exists. But $2\neq3=f(1)$, so condition 3 fails: removable — redefining $f(1)=2$ repairs it.
5 $P(2)=2\lceil2\rceil=4$. Just before $t=2$ the fee is $4$; just after, $\lceil t\rceil=3$ and the fee is $6$: $$\lim_{t\to2^-}P(t)=4,\qquad \lim_{t\to2^+}P(t)=6.$$ Condition 2 fails: a jump discontinuity of $2$ OMR. For the driver, leaving one minute after the two-hour mark costs $2$ OMR more — the fee does not change gradually.

VI. Choosing constants

VI
Find the value(s) of the constant(s) that make each function continuous for all real $x$.
  • 1 $f(x)=\left\{\begin{array}{cl} kx-1 & \quad x<4 \\[2pt] x+k+2 & \quad x\ge4 \end{array}\right.$
  • 2 $f(x)=\left\{\begin{array}{cl} \dfrac{x^2-16}{x-4} & \quad x\neq4 \\[6pt] k & \quad x=4 \end{array}\right.$
  • 3 $f(x)=\left\{\begin{array}{cl} x^2-k & \quad x<3 \\[2pt] kx+1 & \quad x\ge3 \end{array}\right.$
  • 4 $f(x)=\left\{\begin{array}{cl} k^2x & \quad x\le2 \\[2pt] x+6 & \quad x>2 \end{array}\right.$
  • 5 Two constants. $f(x)=\left\{\begin{array}{cl} x+1 & \quad x<1 \\[2pt] ax+b & \quad 1\le x<2 \\[2pt] 3x & \quad x\ge2 \end{array}\right.$
Slide $k$ until the pieces join
current $k$—
left value—
right value—
gap—
1 Both pieces are lines, so only $x=4$ matters. Left limit $4k-1$; right limit and value $4+k+2=k+6$: $$4k-1=k+6\ \Longrightarrow\ 3k=7\ \Longrightarrow\ k=\frac73.$$
2 For $x\neq4$, $\dfrac{(x-4)(x+4)}{x-4}=x+4$, so $\displaystyle\lim_{x\to4}f(x)=8$. Continuity needs $k=f(4)=8$.
3 At $x=3$: left limit $9-k$, right limit and value $3k+1$: $$9-k=3k+1\ \Longrightarrow\ 4k=8\ \Longrightarrow\ k=2.$$ Check: both pieces give $7$ at $x=3$.
4 At $x=2$: value and left limit $2k^2$; right limit $2+6=8$: $$2k^2=8\ \Longrightarrow\ k^2=4\ \Longrightarrow\ k=\pm2.$$ Both values work, because only $k^2$ appears in the function.
5 Two junctions give two equations. At $x=1$: $1+1=a+b$. At $x=2$: $2a+b=3(2)=6$. Subtracting, $$a=6-2=4,\qquad b=2-a=-2.$$ So $a=4,\ b=-2$, and the middle piece is $4x-2$, which runs from $(1,2)$ to $(2,6)$.

VII. Intervals of continuity

VII
State the interval(s) over which each function is continuous.
  • 1 $f(x)=\dfrac{x+2}{x^2-9}$
  • 2 $g(x)=\sqrt{5-x}$
  • 3 $h(x)=\dfrac{x}{x^2+4}$
  • 4 $q(x)=\sqrt{x^2-4}$
Pencil test on each part
pencil position $x$—
status—
lifts so far0
1 A rational function is continuous on its domain. $x^2-9=(x-3)(x+3)=0$ at $x=\pm3$, so $f$ is continuous on $(-\infty,-3)$, $(-3,3)$ and $(3,\infty)$.
2 Need $5-x\ge0$, i.e. $x\le5$. Interior points are fine by the root law, and $\displaystyle\lim_{x\to5^-}\sqrt{5-x}=0=g(5)$, so $g$ is continuous from the left at $5$. Answer: $(-\infty,5]$.
3 The denominator $x^2+4\ge4>0$ never vanishes, so the domain is all of $\mathbb{R}$ and $h$ is continuous on $(-\infty,\infty)$.
4 Need $x^2-4\ge0$, i.e. $x\le-2$ or $x\ge2$. At $x=-2$ the function is continuous from the left, at $x=2$ from the right. Answer: $(-\infty,-2]$ and $[2,\infty)$. On $(-2,2)$ the function does not exist, so the pencil has nothing to draw there.

VIII. The Intermediate Value Theorem

VIII
Recall the Intermediate Value Theorem: if $f$ is continuous on $[a,b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there is at least one $c$ in $(a,b)$ with $f(c)=N$. In particular, if $f(a)$ and $f(b)$ have opposite signs, $f$ has a zero in $(a,b)$.
  • 1 Show that $x^3+x-1=0$ has a root in $[0,1]$. Then narrow it to an interval of length $\tfrac14$.
  • 2 Show that the equation $\cos x=x$ has a solution in $\left[0,\tfrac{\pi}{2}\right]$.
  • 3 Applied. Water in a kettle is at $20^\circ$C when it is switched on and at $95^\circ$C four minutes later. Assuming temperature varies continuously, show that at some moment it was exactly $60^\circ$C.
  • 4 Conceptual. For $f(x)=\dfrac1x$, $f(-1)=-1<0$ and $f(1)=1>0$, yet $f$ has no zero. Why does this not contradict the theorem?
1 $f(x)=x^3+x-1$ is a polynomial, so it is continuous on $[0,1]$. $f(0)=-1<0$ and $f(1)=1>0$, so by the IVT there is $c\in(0,1)$ with $f(c)=0$. Halving: $f\!\left(\tfrac12\right)=\tfrac18+\tfrac12-1=-\tfrac38<0$ and $f\!\left(\tfrac34\right)=\tfrac{27}{64}+\tfrac34-1=\tfrac{11}{64}>0$, so the root lies in $\left(\tfrac12,\tfrac34\right)$. (Numerically, $c\approx0.6823$.)
2 Rewrite as $h(x)=\cos x-x=0$. $h$ is continuous (a difference of continuous functions). $h(0)=1>0$ and $h\!\left(\tfrac{\pi}{2}\right)=0-\tfrac{\pi}{2}<0$, so by the IVT there is $c\in\left(0,\tfrac{\pi}{2}\right)$ with $\cos c=c$.
3 Let $T(t)$ be the temperature after $t$ minutes; $T$ is continuous on $[0,4]$ with $T(0)=20$ and $T(4)=95$. Since $20<60<95$, the IVT with $N=60$ gives some $c\in(0,4)$ with $T(c)=60$.
4 The IVT needs $f$ continuous on the whole interval $[-1,1]$. But $\dfrac1x$ is undefined at $x=0\in[-1,1]$ (an infinite discontinuity), so the hypothesis fails and the theorem promises nothing. The graph jumps from $-\infty$ to $+\infty$ without crossing zero.
IX
Reading continuity from a graph. Use each graph to answer the questions.
Figure 1.13 — $f$ on $[-4,5]$.
Figure 1.14 — $g(x)=1+\dfrac{1}{x-2}$ on $[-3,6]$, hole at $\left(0,\tfrac12\right)$.
Figure 1.15 — $h$ on $[0,6]$.
  • 1 Figure 1.13. (a) For $a=-2,\ 1,\ 3$ find $\displaystyle\lim_{x\to a^-}f(x)$, $\displaystyle\lim_{x\to a^+}f(x)$ and $f(a)$. (b) At which of these points is $f$ continuous? At each discontinuity, state which condition of the definition fails and classify it (removable, jump or infinite). (c) Is $f$ continuous from the left or from the right at $x=-2$? (d) State the interval(s) on which $f$ is continuous.
  • 2 Figure 1.14. (a) Find $\displaystyle\lim_{x\to0}g(x)$ and $g(0)$. (b) Find $\displaystyle\lim_{x\to2^-}g(x)$ and $\displaystyle\lim_{x\to2^+}g(x)$. (c) Classify each discontinuity of $g$. (d) Which discontinuity can be removed by defining one value of $g$? Give that value. (e) State the interval(s) on which $g$ is continuous.
  • 3 Figure 1.15. (a) Check the three conditions of continuity at $x=2$ and at $x=4$. (b) Classify each discontinuity. (c) At which point is $h$ continuous from one side only? Which side? (d) What single change to one value of $h$ makes $h$ continuous at $x=2$? (e) State the interval(s) on which $h$ is continuous.
Check the three conditions on each graph
from the left—
from the right—
value at the point—
?1. value defined
?2. limit exists
?3. limit = value
1(a) $x=-2$: left limit $2$, right limit $3.5$, $f(-2)=2$. $x=1$: both one-sided limits are $-1$, $f(1)=2$. $x=3$: both one-sided limits are $1$, $f(3)=1$.
1(b)–(d) $f$ is continuous at $x=3$ — the corner does not break continuity. At $x=-2$ the one-sided limits differ, so the limit does not exist (condition 2 fails): a jump. At $x=1$ the limit $-1\neq f(1)=2$ (condition 3 fails): removable. At $x=-2$, $f$ is continuous from the left since $\displaystyle\lim_{x\to-2^-}f(x)=2=f(-2)$. Intervals of continuity: $[-4,-2]$, $(-2,1)$, $(1,5]$.
2 (a) $\displaystyle\lim_{x\to0}g(x)=1+\frac{1}{0-2}=\tfrac12$, but $g(0)$ is undefined (open circle). (b) $\displaystyle\lim_{x\to2^-}g(x)=-\infty$, $\displaystyle\lim_{x\to2^+}g(x)=+\infty$. (c) $x=0$: removable (condition 1 fails, the limit exists); $x=2$: infinite (vertical asymptote). (d) Define $g(0)=\tfrac12$; the infinite discontinuity cannot be removed. (e) $[-3,0)$, $(0,2)$, $(2,6]$.
3 (a) $x=2$: $h(2)=4$ is defined, $\displaystyle\lim_{x\to2}h(x)=2$ exists, but $2\neq4$ (condition 3 fails). $x=4$: $h(4)=3$ is defined, but the left limit is $0$ and the right limit is $3$, so the limit does not exist (condition 2 fails). (b) $x=2$ removable, $x=4$ jump. (c) At $x=4$, from the right: $\displaystyle\lim_{x\to4^+}h(x)=3=h(4)$. (d) Redefine $h(2)=2$. (e) $[0,2)$, $(2,4)$, $[4,6]$.
Pattern On a graph: an open circle with no filled dot above or below it means condition 1 fails; a vertical break between the two sides means condition 2 fails; a filled dot away from where both sides meet means condition 3 fails. Corners are still continuous.
🔬 Research–Teaching NexusResearch-based · optional, about 30 minutes

Research question. Which everyday prices in Nizwa jump suddenly, and what type of discontinuity does each jump create?

Why it matters. Economists and transport planners model tariffs with piecewise functions; jumps at thresholds change how people behave.

What to do.

  1. Choose one: a taxi fare, a mobile data package, or a parking fee. Find the real rules (receipt, operator website, sign) or use this sample: parking costs 200 baisa for the first hour, then 100 baisa for each extra hour or part of an hour.
  2. Write the cost $C(t)$ as a piecewise function and sketch it.
  3. List every point where $C$ is not continuous. Compute the left and right limits there.
  4. Classify each discontinuity (jump, removable, infinite).

Tools: Graph paper or Desmos

Hand in: A half-page note: the rule, the graph, and a table of discontinuities with their type.

Going further. Anywhere: compare two providers' tariffs and say which one has fewer jumps for a typical user.

Instructor note: Students should notice that “part of an hour” gives a step (ceiling) function with jump discontinuities, and decide whether the function is left- or right-continuous at each step.

🧪 Self-check quiz

0/5 correct

Q1. Which condition fails for $f(x)=\dfrac{x^2-1}{x-1}$ at $x=1$?

Q2. $f(x)=\dfrac{x+1}{x-5}$ is continuous on:

Q3. Why is $\sqrt{4-x^2}$ continuous on the closed interval $[-2,2]$?

Q4. For $g(x)=\left\{\begin{array}{cl} \sqrt{kx} & \quad 0\le x\le3 \\[2pt] x+1 & \quad 3\le x\le10 \end{array}\right.$ the value of $k$ giving continuity is:

Q5. A function has $\displaystyle\lim_{x\to a^-}f(x)=5$, $\displaystyle\lim_{x\to a^+}f(x)=5$ and $f(a)=2$. The discontinuity at $a$ is:

← 1.2 The Limit Laws1.4 Limits Involving Infinity →