NCEA Level 3Continuity and differentiabilityAS 91578

Limits, continuity and differentiability from a graph

What this lesson covers

A function is continuous where its graph has no break, and differentiable where it also has no sharp corner.
Reading these from a piecewise graph is Achieved-level work that is often left blank.

Key idea

Three questions about a point

Does the limit exist?
Is the function continuous there?
Is it differentiable there?

Worked example

A graph in three pieces

A curve, maximum at (−1, 3).
Open dot at (1, −1), closed dot at (1, 2).
Up to (3, 4), then down to (5, 0).

Key idea

Open and closed dots

Open: not on the graph.
Closed: on the graph.
f(1) = 2, not −1.


Open dot	not on the graph
Closed dot	the value
(1, −1)	open, not included
f(1)	2, the closed dot

Worked example

The limit as x tends to 1

From the left: heading to −1.
From the right: heading to 2.
They disagree: no limit at x = 1.

Worked example

Writing the limits

Left −1, right 2: no limit.
Where it heads, not f(1).

from the left: f(x) → −1as x → 1
from the right: f(x) → 2as x → 1
−1 ≠ 2the sides disagree
lim(x→1) f(x) does not exist
x → 3, both sides: f(x) → 4
lim(x→3) f(x) = 4

Worked example

Continuous: no break

Draw through it without lifting the pen.
The limit exists and equals the value.
x = 1: a jump, not continuous.
x = 3: continuous, with a corner.

Worked example

Differentiable: one clear gradient

Continuous first.
Then one gradient.
A corner has two.

x = 1: a jump, not continuous
not differentiable at x = 1
x = 3: lim f(x) = 4 = f(3)continuous
left: (4 − 2)/(3 − 1) = 1rise over run
right: (0 − 4)/(5 − 3) = −2rise over run
1 ≠ −2: a cornergradients differ
not differentiable at x = 3

Worked example

Easy marks to lose

The open dot is not the value.
A jump has no limit.
Not differentiable: the jump too.

	Misread	Right
f(1)	−1	2, closed dot
Limit at 1	2	does not exist
Differentiable at 3	yes, no gap	no, a corner
Not differentiable	x = 3 only	x = 1 and x = 3

Worked example

The sign of the gradient

Rising: positive. Falling: negative.
Zero at the maximum, x = −1.
The curve is concave down.

	f'(x)
−3 < x < −1	positive
x = −1	zero, maximum
−1 < x < 1	negative
1 < x < 3	positive, 1
3 < x < 5	negative, −2

Worked example

Unknown constants

Pieces meet at x = 1.
Gradients match at x = 1.

f(x) = ax² + b (x < 1), 4x (x ≥ 1)
a(1)² + b = 4(1)pieces meet
a + b = 4
f'(x): 2ax (x < 1), 4 (x > 1)
2a(1) = 4, so a = 2gradients match
b = 4 − 2 = 2substitute back
a = 2, b = 2
2(1)² + 2 = 4 = 4(1)

Key idea

Where the marks are

Achieved: read them from a graph.
Merit: explain each, with the reason.
Excellence: combined with other calculus.

Summary

What to take away

Read the dots, then check both sides, then look for jumps and corners.


Open dot	not on the graph
Closed dot	the value
Limit	both sides agree
Continuous	limit equals value
Differentiable	no break, corner, cusp
Gradient sign	rising or falling

Your turn

Your turn: four questions

Pause here and answer each one.
Question 3: the absolute value of x − 3.
Give a reason with every answer.

	Function	Find
1	x+2 (x<1), 5−x (x≥1)	continuous at 1?
2	x² (x<2), 4x−4 (x≥2)	differentiable at 2?
3	\|x − 3\|	not differentiable at
4	kx+1 (x<2), x² (x≥2)	k for continuity

Show answer

1. Left limit 3, right limit 4: a jump, so not continuous at x = 1. 2. Both pieces give 4 and both gradients are 4: differentiable at x = 2. 3. x = 3: a corner, gradient −1 on the left and 1 on the right. 4. 2k + 1 = 4, so k = 3/2.

Worked example

The answers: questions 1 and 2

1. left: 1 + 2 = 3x + 2 as x → 1
right: f(1) = 5 − 1 = 45 − x at x = 1
3 ≠ 4: a jump, not continuous
2. 2² = 4 and 4(2) − 4 = 4pieces meet
gradients: 2x = 2(2) = 4, and 4
they match: differentiable at 2

Worked example

The answers: questions 3 and 4

3. |x − 3| = 3 − x for x < 3
|x − 3| = x − 3 for x ≥ 3
gradients −1 and 1 at x = 3
a corner: not differentiable
4. k(2) + 1 = 2²pieces meet at x = 2
2k + 1 = 4, so 2k = 3
k = 3/2

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