A. Rate of Change · Limits and Rates of Change

Limit Laws and Continuity

A1.4 A1.2About 70 minutes

Learning goals

I can use the limit laws to evaluate limits, and explain when direct substitution works.
I can state the three conditions for a function to be continuous at a point.
I can classify discontinuities as removable, jump or infinite.
I can find a constant that makes a piecewise function continuous.
I can describe continuity in real situations such as fees and fares.

Words to know

limit laws
Rules for limits of sums, differences, multiples, products, quotients, powers and roots, such as lim (f + g) = lim f + lim g.

direct substitution
Finding lim (x → a) f(x) by calculating f(a). It works whenever f is continuous at a.

continuous at a
f is continuous at x = a if f(a) is defined, lim (x → a) f(x) exists, and the two are equal.

removable discontinuity
A hole: the limit exists but f(a) is undefined or different from it. Redefining one point removes it.

jump discontinuity
The left-hand and right-hand limits exist but are different.

infinite discontinuity
f(x) grows without bound near a (a vertical asymptote).

Shortcuts for limits

Tables and graphs are great for discovering limits, but slow. The limit laws let you compute most limits directly.

Suppose lim (x → a) f(x) = L and lim (x → a) g(x) = M, where L and M are real numbers. Then, as x → a:

Sum and difference: lim (f ± g) = L ± M
Constant multiple: lim (c·f) = c·L
Product: lim (f·g) = L·M
Quotient: lim (f/g) = L/M, provided M ≠ 0
Power: lim fⁿ = Lⁿ
Root: lim ⁿ√f = ⁿ√L (for even n, L must be at least 0)

Two basic facts start everything off: lim (x → a) c = c, and lim (x → a) x = a.

Direct substitution

Put the laws together and you get a powerful result. For any polynomial p,

lim (x → a) p(x) = p(a).

For a rational function p/q, lim (x → a) p(x)/q(x) = p(a)/q(a) as long as q(a) ≠ 0. The same is true for roots, exponential functions and sine and cosine at every point of their domains.

So the first thing to try with any limit is to substitute. If you get a real number, you are done. If you get 0/0, the next two lessons show what to do. If you get (non-zero)/0, there is a vertical asymptote (last lesson).

Worked example

Example 1: Using the limit laws

Evaluate lim (x → 2) (3x² − 5x + 4)/(x + 1), justifying each step.

Denominator: lim (x + 1) = 2 + 1 = 3 ≠ 0, so the quotient law applies.
Numerator: 3·lim x² − 5·lim x + lim 4 = 3(2)² − 5(2) + 4 = 12 − 10 + 4 = 6 (constant multiple, power, difference and sum laws).
Quotient law: 6 ÷ 3 = 2.

2. In practice you just substitute x = 2, but the laws are the reason it works.

Worked example

Example 2: Laws with unknown functions

Given lim (x → 5) f(x) = 4 and lim (x → 5) g(x) = −2, evaluate (a) lim [3f(x) − g(x)²], (b) lim f(x)/g(x), (c) lim √f(x), (d) lim [f(x) + x]·g(x).

(a) 3(4) − (−2)² = 12 − 4 = 8.
(b) 4 ÷ (−2) = −2 (allowed because −2 ≠ 0).
(c) √4 = 2.
(d) (4 + 5)(−2) = −18. Remember lim (x → 5) x = 5.

(a) 8, (b) −2, (c) 2, (d) −18

Continuity

Informally, a function is continuous if you can draw its graph without lifting your pencil. Here is the precise version.

f is continuous at x = a if all three are true:

f(a) is defined;
lim (x → a) f(x) exists;
lim (x → a) f(x) = f(a).

If any condition fails, f is discontinuous at a. There are three common types:

Removable: the limit exists, but f(a) is missing or is the wrong value (a hole). You could fix it by redefining one point.
Jump: the one-sided limits exist but are different.
Infinite: f(x) is unbounded near a (a vertical asymptote).

Polynomials, sine and cosine, and exponential functions are continuous everywhere. Rational and root functions are continuous everywhere on their domains. A smooth function, as the curriculum describes it, is continuous and has no corners.

Three kinds of discontinuity at x = a. Open dots are points not on the graph; filled dots are actual values.

Worked example

Example 3: Classifying discontinuities

Where is f(x) = (x² − 1)/(x² − 3x + 2) discontinuous? Classify each discontinuity.

The denominator is 0 when x² − 3x + 2 = (x − 1)(x − 2) = 0, so f is undefined at x = 1 and x = 2.
Factor the numerator too: f(x) = (x − 1)(x + 1)/((x − 1)(x − 2)) = (x + 1)/(x − 2) for x ≠ 1.
At x = 1: the limit is (1 + 1)/(1 − 2) = −2. The limit exists but f(1) is undefined: removable (a hole at (1, −2)).
At x = 2: the simplified numerator → 3 while the denominator → 0, so f is unbounded: infinite discontinuity (a vertical asymptote).

Removable discontinuity at x = 1 (hole at (1, −2)); infinite discontinuity at x = 2.

Worked example

Example 4: Making a piecewise function continuous

f(x) = x² + k for x < 2 and f(x) = 3x − 1 for x ≥ 2. Find k so that f is continuous everywhere.

Each piece is a polynomial, so the only possible problem is at x = 2.
f(2) = 3(2) − 1 = 5, and the right-hand limit is also 5.
Left-hand limit: 2² + k = 4 + k.
For continuity: 4 + k = 5, so k = 1.

k = 1. Then both pieces meet at (2, 5).

Worked example

Example 5: A discontinuity in real life

A parking lot charges $4 for each hour or part of an hour. Write the one-sided limits of the cost C at t = 2 h. Is C continuous at t = 2?

For t a little under 2 h (say 1 h 59 min), you pay for 2 hours: $8. So lim (t → 2⁻) C(t) = 8.
Parking exactly 2 h also costs $8, so C(2) = 8.
For t a little over 2 h (2 h 1 min), you pay for 3 hours: $12. So lim (t → 2⁺) C(t) = 12.
The one-sided limits differ.

C has a jump discontinuity at t = 2 (and at every whole number of hours). It is continuous on each open interval between whole hours.

The parking cost is a step function. At each whole hour the cost jumps by $4, so the function is discontinuous there.

Just for fun

The parking lot's step function may be the only graph that can make you sprint across a parking lot. Arriving at 2 hours and 1 minute costs you a whole extra step.

Hands-on activity

Find the jumps in your world

You need: a pencil, grid paper, and some real price lists (a postage rate chart, a phone or data plan, a parking sign, a taxi or ride-share fare, or income tax brackets from a government website).

Choose two price rules. Write each one as a function of a quantity (mass, minutes, kilometres, income).
Graph each rule on grid paper for a sensible range.
Mark every discontinuity. For each one, write the left-hand limit, the right-hand limit and the actual value (use open and filled dots).
Find one quantity in daily life that changes continuously (for example, the temperature outside, or the water level in a bathtub) and explain why its graph has no jumps.

What to look for: mailing and parking rules usually give step functions with jump discontinuities at each threshold. A per-kilogram price (like bulk food) is continuous. Temperatures and water levels change continuously because they cannot skip values: to go from 2 °C to 5 °C, the temperature passes through every value in between.

Your turn

Evaluate lim (x → 3) (x³ − 2x + 1).

Evaluate lim (x → −1) (x² + 4)/(x − 2).

Given lim (x → a) f(x) = 4 and lim (x → a) g(x) = −2, find lim [3f(x) − g(x)²].

With the same f and g, find lim [f(x)·g(x)] and lim √f(x).

Is f(x) = 1/(x − 5) continuous at x = 5? Classify.

f(x) = (x² − 9)/(x − 3). Classify the discontinuity at x = 3 and say how to remove it.

f(x) = kx + 1 for x < 1 and f(x) = x² + 3 for x ≥ 1. Find k so f is continuous.

More practice: warm-up

Evaluate lim (x → 2) (x + 5).

Evaluate lim (x → 0) cos x.

List the three conditions for f to be continuous at x = a.

Is every polynomial continuous everywhere?

A graph has a single open dot (a hole) at x = 4 and no other breaks. What type of discontinuity is this?

More practice: on your own

Evaluate lim (x → 1) (2x + 3)⁴.

Evaluate lim (x → 8) (∛x + x^(2/3)).

Where is f(x) = (x + 2)/(x² − 4) discontinuous? Classify each discontinuity.

Using the parking lot ($4 per hour or part), find C(2.5), lim (t → 3⁻) C(t) and lim (t → 3⁺) C(t).

f(x) = x + a for x < 1, f(1) = 3, and f(x) = bx² for x > 1. Find a and b so f is continuous at 1.

Challenge

f(x) = (x² + ax + 6)/(x − 2). Find a so that lim (x → 2) f(x) exists, then find the limit.

The Intermediate Value Theorem says: if f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c in (a, b). Use it to show x³ + x − 1 = 0 has a root between 0 and 1.

Give functions f and g so that lim (x → 0) f(x) and lim (x → 0) g(x) do not exist, but lim (x → 0) [f(x) + g(x)] does. Why doesn't this break the sum law?

Check your understanding

Quiz

Evaluate lim (x → 2) (x³ − 4x + 7).

If lim (x → a) f(x) = 6 and lim (x → a) g(x) = 3, what is lim (x → a) [f(x)/g(x) − g(x)]?

Which condition is not required for f to be continuous at x = a?

f(x) = (x² − 4)/(x − 2). What kind of discontinuity is at x = 2?

f(x) = 2x + k for x < 3 and f(x) = x² for x ≥ 3. Which k makes f continuous?

A parking lot charges $5 per hour or part of an hour. At t = 1 h, the cost function has:

For parents and teachers

Big idea (A1.4, A1.2): the limit laws justify direct substitution for continuous functions. A function is continuous at a when f(a), the limit, and their equality all check out. Discontinuities are removable, jump or infinite. The curriculum's "smooth" functions are continuous with no corners.

How to help: Look at real price rules together (postage, parking, data plans). Ask your teen where the jumps are and what the one-sided limits are at each jump.

Common difficulties:

Using the quotient law when the denominator's limit is 0.
Calling a hole an asymptote, or the reverse. Factor first: a factor that cancels gives a hole.
Checking only that pieces of a piecewise function exist, not that they meet.
Thinking continuous means smooth. |x| is continuous everywhere but not smooth at 0.