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Grade 12 Calculus · Lesson 1.490 min

Continuity & One-Sided Limits

See continuity, diagnose every failure, make piecewise functions meet, and use IVT and bisection to trap a zero.

Test continuity at a point, classify discontinuities, analyze piecewise and special functions, check closed intervals, and apply the Intermediate Value Theorem.

3 continuity checksDiscontinuitiesPiecewiseAbsolute value & GIFClosed intervalsIVTBisection
01

Start with the definition

Continuity at a Point

Interactive lab

The three continuity checks

Inspect the evidence at c = 2. Predict the first condition that fails, or decide that all three hold. Changing the case starts a new prediction.
0011.52334.546xf(x)
Case: continuous. At c = 2, left limit 3, right limit 3, function value 3. Axes use independent scales. Open circles exclude a point; filled circles include it.

Function value

f(c) = 3

Left limit

L⁻ = 3

Right limit

L⁺ = 3
Which continuity check fails first?
Explanation

A function is continuous at x=cx=c only when all three conditions are true.

1 · Function valuef(c) is definedf(c)\text{ is defined}
2 · Limit existslim⁡x→c−f(x)=lim⁡x→c+f(x)\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)
3 · They matchlim⁡x→cf(x)=f(c)\lim_{x\to c}f(x)=f(c)

Your turn

For ff to be continuous at x=cx=c, which statement must be true?
02

Identify how continuity fails

Types of Discontinuity

Explanation

Removable

The two-sided limit exists, but the point is missing or has the wrong value.

Explanation

Jump

Finite one-sided limits approach different values.

Explanation

Infinite

The function grows without bound near a vertical asymptote.

Worked example · removable discontinuity

f(x)=x2−1x−1,x=1f(x)=\frac{x^2-1}{x-1},\quad x=1

Use the three checks before classifying the discontinuity.

Predict before you reveal

What value repairs the missing point at x=1x=1?
Common mistake. A filled point does not guarantee continuity. The limit must exist and equal the function value.

Your turn

If lim⁡x→a−f(x)\lim_{x\to a^-}f(x) and lim⁡x→a+f(x)\lim_{x\to a^+}f(x) are finite but unequal, what type of discontinuity occurs at x=ax=a?
03

Use the family before doing extra work

Where Common Functions Are Continuous

Explanation

Continuous for all real numbers

Polynomials · absolute value of polynomial expressions · sine and cosine of polynomial expressions.

Explanation

Continuous on their domains

Rational functions: denominator ≠ 0. Even-index radicals: respect the real domain. Odd-index radicals: continuous wherever defined.

Key takeawayMr. Azzam’s shortcut

Find the domain first. Then focus on excluded values, piecewise breaking points, and step-function points.

Your turn

Where can f(x)=x+1x2−9f(x)=\dfrac{x+1}{x^2-9} fail to be continuous?
04

Left rule · right rule · function value

Piecewise Continuity at Breaking Points

Interactive lab

Make the pieces meet

Adjust k to make this piecewise function continuous at x = 2. Use the one-sided values to test your choice, then check your prediction.
g(x)={x2+k,x<23x−1,x≥2g(x)=\begin{cases}x^2+k,&x<2\\3x-1,&x\ge2\end{cases}
0-30.7501.532.25639xf(x)
Piecewise function with k = 0; the left endpoint is open at (2, 4) and the value is filled at (2, 5). Axes use independent scales. Open circles exclude a point; filled circles include it.

Left limit

4 + k = 4

Right limit and g(2)

3(2) − 1 = 5
Is the current choice continuous at 2?
lim⁡x→c−f(x)=lim⁡x→c+f(x)=f(c)\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)

Worked example · a different parameter

h(x)={2x+m,x<3x2−1,x≥3h(x)=\begin{cases}2x+m,&x<3\\x^2-1,&x\ge3\end{cases}

Find m so that h is continuous at x=3.

Predict before you reveal

Which continuity equation determines mm?

Your turn

At a breaking point x=cx=c of a piecewise function, what should you compare?
05

Rewrite special functions first

Absolute Value & Greatest Integer Functions

Explanation

Absolute value

Find where the inside equals zero, rewrite piecewise, then test the breaking point.

∣x+7∣={−(x+7),x<−7x+7,x≥−7|x+7|=\begin{cases}-(x+7),&x<-7\\x+7,&x\ge-7\end{cases}

Worked example · different absolute value

g(x)=∣2x−6∣,x=3g(x)=|2x-6|,\quad x=3

Predict the outcome, then inspect each worked step.

Predict before you reveal

At x=3x=3, what is the common limit and value of ∣2x−6∣|2x-6|?
Explanation

Greatest integer parent function

f(x)=⌊x⌋,lim⁡x→n−⌊x⌋=n−1≠n=lim⁡x→n+⌊x⌋=f(n)f(x)=\lfloor x\rfloor,\qquad \lim_{x\to n^-}\lfloor x\rfloor=n-1\ne n=\lim_{x\to n^+}\lfloor x\rfloor=f(n)

Therefore the parent greatest integer function has a jump at every integer.

Worked example · transformed greatest integer

f(x)=x⌊x⌋,x=0f(x)=x\lfloor x\rfloor,\quad x=0

Do not assume the transformation is discontinuous. Test it.

Predict before you reveal

Is x⌊x⌋x\lfloor x\rfloor continuous at 00?
Common mistake. Containing a greatest integer function does not automatically make every transformed function discontinuous at every integer.

Your turn

For f(x)=⌊x⌋f(x)=\lfloor x\rfloor, compare the one-sided limits at an integer nn.
06

Endpoints use one side

Continuity on a Closed Interval

Explanation

Use two-sided continuity inside the interval and the side that lies inside the interval at each endpoint.

continuous on (a,b),lim⁡x→a+f(x)=f(a),lim⁡x→b−f(x)=f(b)\text{continuous on }(a,b),\qquad\lim_{x\to a^+}f(x)=f(a),\qquad\lim_{x\to b^-}f(x)=f(b)

Your turn

To be continuous on [a,b][a,b], what is required at aa?
07

A continuous curve cannot skip an intermediate level

Intermediate Value Theorem

Interactive lab

Intermediate values: check the guarantee

f(x)=x3,−1≤x≤1f(x)=x^3,\quad -1\le x\le1. This polynomial is continuous. Move the target K, then decide whether the endpoint values guarantee a solution in the closed interval.
-1-1.5-0.5-0.75000.50.7511.5xf(x)K
f(x) = x cubed on [−1, 1]; target K = 0.0. Axes use independent scales. Open circles exclude a point; filled circles include it.

Left endpoint

f(−1) = −1

Right endpoint

f(1) = 1

Target

K = 0.0
Does IVT guarantee this level on [−1, 1]?
Definition

IVT

If ff is continuous on [a,b][a,b] and KK lies strictly between f(a)f(a) and f(b)f(b), then at least one c∈(a,b)c\in(a,b) satisfies f(c)=Kf(c)=K.

Worked example · prove a different zero exists

g(x)=x3+2x−2on [0,1]g(x)=x^3+2x-2\quad\text{on }[0,1]

Check the two facts IVT needs before making the conclusion.

Predict before you reveal

What does IVT alone prove for this polynomial on [0,1][0,1]?
Common mistake. IVT guarantees at least one solution; it does not say the solution is unique.

Your turn

If ff is continuous on [a,b][a,b] and KK lies between f(a)f(a) and f(b)f(b), IVT guarantees:
08

Halve · test signs · keep the bracket

Bisection Method · Zero Approximation

Interactive lab

Trap the zero with successive brackets

f(x)=x3+2x−1f(x)=x^3+2x-1 is continuous. Starting with [0, 1], use the signs at a, m and b to choose the half that still traps a zero. Each correct choice creates the next interval.

Left endpoint

a = 0
f(a) = -1

Midpoint

m = 0.5
f(m) = 0.125

Right endpoint

b = 1
f(b) = 2
0-10.25-0.250.50.50.751.2512xf(x)0
Current bracket [0, 1]. Root crossings occur at output zero. Axes use independent scales. Open circles exclude a point; filled circles include it.

Midpoint estimate: 0.5. Guaranteed absolute error ≤ 0.5. Display values are rounded; interval decisions use full precision.

Which half retains opposite endpoint signs?

Review bracket history (0 steps)
Bracket history · 0 successful bisections
StepabMidpoint error bound
0010.5
Key takeaway

STEP 1

Bracket

Opposite signs

STEP 2

Midpoint

m=(a+b)/2m=(a+b)/2

STEP 3

Evaluate

f(m)f(m)

STEP 4

Repeat

Keep sign-change half

Worked example · bisection on a different function

g(x)=x3−x−1,[1,2]g(x)=x^3-x-1,\quad[1,2]

Choose the half-interval that keeps opposite signs.

Predict before you reveal

After the first midpoint test, which interval retains the sign change?

Your turn

For a continuous function with opposite endpoint signs, the midpoint output is nonzero. Which half-interval should be kept?

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