MAT01A1: Limitse-Quiz 3 I Opens Thursday afternoon. I Closes Friday 9 March at 23h59. I Covers...
Transcript of MAT01A1: Limitse-Quiz 3 I Opens Thursday afternoon. I Closes Friday 9 March at 23h59. I Covers...
MAT01A1: Limits
Dr Craig
28 February 2018
Note: these lecture slides should be inused in conjunction with the textbook.These slides do not tell the full story.You must read the sections from thetextbook and complete the prescribedtutorial exercises in order to fullyunderstand each section.
Semester Test 1
I Saturday 10 March at 08h30
I See detailed announcement on
Blackboard
Saturday class this week
I 09h00 – 12h00
I C-LES 402
e-Quiz 3
I Opens Thursday afternoon.
I Closes Friday 9 March at 23h59.
I Covers Chapter 1.
Tut 5
Questions are available under “Tutorials”.
Past paper
Semester Test 1 from 2017 available under
“Assessments”.
Some inverse trig examples
Calculate the following
I arcsin(−√3
2
)I cos(arcsin t)
I cos(arctan 23)
I Tough one:
sec(arcsin(2x− 1))
Today’s lecture: Limits
I Why limits?
I The definition of a limit
I Estimating limits by calculation
I One-sided limits
I Infinite limits
I Vertical asymptotes
I Khan Academy exercises on limits
The tangent to a function
Consider the parabola
y = x2
We are interested in the tangent to f (x) at
the point P = (1, 1).
We take different values of Q (with Q 6= 1)
and calculate the gradient of the straight line
that goes through P and Q.
Values of mPQ
x < 1 mPQ
0 1
0.5 1.5
0.9 1.9
0.99 1.99
0.999 1.999
x > 1 f (x)
2 3
1.5 2.5
1.1 2.1
1.01 2.01
1.001 2.001
The tangent line as Q approaches P
Playing around with limits
Go to the Stewart Calculus link:
Tools for enriching calculus
Click on “Limits and Derivatives”, then click
on “Secant Line and Tangent Line”. The red
dot (x = a) represents the fixed point P and
by adjusting the value of h you can move the
blue dot Q.
Velocity
We can easily measure distance with respect
to time, but to measure instantaneous
velocity we need to use limits.
Dropping a ball from a tower
Definition of a limit of a function
Definition: Suppose f(x) is defined for x near*a. If we can get the values of f(x) as close to Las we like by taking x as close to a on either side(but not equal to a), then we write
limx→a
f(x) = Land say“the limit of f(x), as x approaches a, equals L”
*near: f(x) is defined on some open intervalincluding a, but possibly not at a itself.
Example
Estimate the value of limx→1
x− 1
x2 − 1.
x < 1 f(x)
0.5 0.666667
0.9 0.526316
0.99 0.502513
0.999 0.500250
0.9999 0.500025
x > 1 f(x)
1.5 0.4
1.1 0.476190
1.01 0.497512
1.001 0.499750
1.0001 0.499975
Based on the above calculations, we estimate that
limx→1
x− 1
x2 − 1=
1
2.
Example:
Consider the function f (x) =x2 + 4x + 4
x + 2.
Observe that
f (x) =(x + 2)(x + 2)
x + 2
Is this different from the function
g(x) = x + 2?
Example
g(x) =
{x + 3 if x 6= 2
1 if x = 2
Sketch the function and find limx→2
g(x).
One-sided limits
Definition: We write
limx→a−
f(x) = L
and say “the limit of f(x) as x approaches afrom the left” is equal to L if we can make thevalues of f(x) as close as we like to L by takingx close to a, but with x less than a.
This is also called “the left-hand limit of f(x) as xapproaches a”
From the other side
Definition: We write
limx→a+
f(x) = L
and say “the limit of f(x) as x approaches afrom the right” is equal to L if we can make thevalues of f(x) as close as we like to L by takingx close to a, but with x greater than a.
This is also called “the right-hand limit of f(x) as xapproaches a”
One-sided example
h(x) =
cosx if x < 012 if x = 0
x if x > 0
limx→0−
h(x) = 1, limx→0+
h(x) = 0, h(0) =1
2
One (sided) + one (sided) = two (sided)
For L ∈ R we have the following:
limx→a
f (x) = L if and only if
limx→a−
f (x) = L and limx→a+
f (x) = L
If there exists L ∈ R such that
limx→a
f (x) = L then we say:
“the limit of f (x) as x approaches a exists”
One (sided) + one (sided) = two (sided)
In the previous two examples of piece-wise
functions we have:
I limx→2
g(x) exists
(since limx→2−
g(x) = 5 = limx→2+
g(x))
I limx→0
h(x) does not exist
(because limx→0−
h(x) 6= limx→0+
h(x))
Infinite limits:
Definition: Let f be a function defined on
both sides of a, except possibly at a itself.
Then
limx→a
f (x) =∞
means that f (x) can get as big as we like
by taking x sufficiently close to a.
Example: limx→0
1
x2=∞
One-sided infinite limits:
Let f (x) =1
x,
and consider
limx→0−
f (x) and limx→0+
f (x)
Let g(x) = cscx, and consider
limx→π−
g(x) and limx→π+
g(x)
Very important:
For the limit at x approaches a of f (x) to
exist, we need
limx→a
f (x) = L
for some L ∈ R. Therefore, limx→0
1
x2does not
exist.
Example: find the following limits:
limx→3+
2x
x− 3and lim
x→3−
2x
x− 3
To calculate the first limit, consider an x
value just a little bit bigger than 3 (e.g. 3.1)
and then see what happens as you get closer
to 3 (e.g. 3.01).
To calculate the second limit, consider an x
value slightly smaller than 3 (e.g. 2.9) and
then get closer to 3 (e.g. 2.99).
Vertical asymptotes
The line x = a is called a verticalasymptote of the curve y = f (x) if at
least one of the following statements is
true:
limx→a
f (x) =∞ limx→a
f (x) = −∞
limx→a−
f (x) =∞ limx→a−
f (x) = −∞
limx→a+
f (x) =∞ limx→a+
f (x) = −∞
Important: as stated on the previous slide, youonly need one of the listed conditions to be satisfiedin order to have a vertical asymptote at x = a; e.g.:
f(x) =
{1 if x 6 0
lnx if x > 0
limx→0+
f(x) = −∞ so f(x) has a V.A. at x = 0.
Example
Find the vertical asymptotes of the following
function:
f (x) =x2 + 2
4x + 4x2
You should find two x values where f (x)
might have a vertical asymptote. Call these
values a and b. Now find
limx→a+
f (x) and limx→a−
f (x)
and
limx→b+
f (x) and limx→b−
f (x)
Khan Academy exercises on limits
Click on the hyperlinks to test your
understanding of limits:
One-sided limits from graphs
Two-sided limits from graphs