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# MATH 1003 Calculus and Linear Algebra

(Lecture 13)

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Limits

Example
x2 1
Graph the function f (x) = . Then find out how f (x)
x +1
behaves as x tends to, but not equal to 1.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Limits

Remarks
I As x gets close to 1, f (x) gets close to 2.
I We say that the limit of f (x) is 2 as x tends to 1.
I Notice that f (x) is undefined when x = 1.
I The limit of f (x) as x tends to 1 is well-defined even when
f (x) is undefined at 1. In fact, the value of f at 1 does
not affect the limit.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Definition of the Limit of a Function

Definition
We write

## lim f (x) = L or f (x) L as x c

xc

if f (x) is close to the single real number L when x tends to, but
not equal to c.

Remarks
I Limit does not always exist i.e. it may happen that f (x) does
not get close to any real number as x tends to a certain c.
I Limit is unique if it exists.

## The limit of a function f (x) at c is determined by the evaluation

of f (x) in the neighbourhood of c.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

One-sided Limits

Example
|x|
Let h(x) = . Find h(0) and lim h(x).
x x0

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

One-sided Limits

Remarks
I As x gets close to 0 from the right hand side, f (x) gets close
to 1.
I As x gets close to 0 from the left hand side, f (x) gets close to
1.
I If x gets close to 0 without restricting the direction of
approach, f (x) cannot be close to a single value.

## We saw that the values of the function h(x) approached two

different numbers, depending on the direction of approach, and it
was natural to refer to these value as the limit from the left and
the limit from the right.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Left-hand and Right-hand Limits
Example
Find lim+ h(x), lim h(x) and lim h(x).
x0 x0 x0

Solution
I lim h(x) = 1.
x0+
I lim h(x) = 1.
x0
I lim h(x) does not exist.
x0

Theorem
A limit exists means that the left-hand limit and the right-hand
limit must be equal. That is,

xc xc xc +

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Finding limits without graph

Example
(x + 1)2 1
Find lim f (x) = lim .
x0 x0 x
Solution
The idea is to evaluate f (x) for various values of x which are close
to 0 e.g. x = 0.01, 0.001, 0.01, 0.001 and so on to guess the
limit. Consider the following table:

## x 0.01 0.001 0.0001 0.0001 0.001 0.01

f (x) 1.99 1.999 1.9999 2.0001 2.001 2.01

## Notice that f (x) tends to 2 as x tends to 0. Therefore,

(x + 1)2 1
lim f (x) = lim =2
x0 x0 x

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Finding limits without graph

Remarks
I In the previous example, f (x) can in fact be simplified in order
to make the calculation easier:
(x + 1)2 1 x 2 + 2x
f (x) = = =x +2
x x
whenever x 6= 0. Therefore, it is clear that lim f (x) = 2.
x0
I When evaluating one-sided limit, say lim+ f (x), the values of
x0
x we use must be on one side i.e. x = 0.01, 0.001 and so on.
(Note: for lim f (x), we use x = 0.01, 0.001 and so on)
x0

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Average Rate of Change

## Let y = f (x). Roughly speaking, the rate of change of y is

Change in y
Change in x
More rigorously, we have the following definition:
Definition
For y = f (x), the average rate of change from x = a to x = b is

f (b) f (a)
ba

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Average Rate of Change

Example
A small ball dropped from a tower will fall a distance of y feet in x
seconds, as given by the formula

y = 16x 2 .

## (a) Find the average velocity from x = 2 seconds to x = 3

seconds.
(b) Find the average velocity from x = 2 seconds to x = 2 + h
seconds, h 6= 0.
(c) Find the expression from part (2) as h 0, if it exists.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Average Rate of Change

Solution
16 32 16 22
(a) The average velocity is = 80 feet per
32
second.
16(2 + h)2 16 22 16(h2 + 4h)
(b) The average velocity is =
2+h2 h
feet per second.
16(h2 + 4h)
(c) lim = lim 16(h + 4) = 64 feet per second.
h0 h h0

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Instantaneous Rate of Change

## In the previous example, we consider the average rate of change of

distance when the change of x is from 2 to 2 + h. And then we let
h tends to 0. The limit can be regarded as the instantaneous rate
of change of at 2. In general, we have the following definition:
Definition
For y = f (x), the instantaneous rate of change at x = a is

f (a + h) f (a)
lim
h0 h
i.e. it is the limit of the difference quotient of f at x = a.

Example
The instantaneous rate of change at x = 2 in the previous example
is 64.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Slope of a Secant Line

## A line through two point on the graph

of y = f (x) is called a secant line. If
(a, f (a) and (a + h, f (a + h)) are two
points on the graph of y = f (x), then
the slope of secant line from x = a to
x = a + h is
f (a + h) f (a) f (a + h) f (a)
= .
(a + h) a h

## Thus, the slope of secant line can be in-

terpreted as the average rate of change
of y from x = a to x = a + h.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Slope of a Secant Line

Example
Given y = f (x) = 0.5x 2 ,
(a) Find the slope of secant line for a = 1, and h = 2.
(b) Find the slope of secant line for a = 1 and h for any nonzero
number.
(c) Find the limit of expression in (b) as h 0.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Slope of a Secant Line

Solution
(a) The slope of secant line is

f (3) f (1)
=2
2
(b) The slope of secant line is

## f (1 + h) f (1) 0.5(1 + h)2 0.5 h + 0.5h2

= = = 1 + 0.5h
h h h
(c) As h 0, we have

f (1 + h) f (1)
lim = lim (1 + 0.5h) = 1
h0 h h0

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Slope of a Tangent

## From the graph, we observe that the

slope of the secant line tends to the slope
of the tangent as h tends to 0. There-
fore, we have the following definition:
Definition
Given y = f (x), the slope of the
tangent line of f(x) at the point x = a
is given by

f (a + h) f (a)
lim
h0 h
if the limit exists.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

The Derivative

Definition
For y = f (x), we define the derivative of f at x, denoted by f 0 (x),
dy df
or , to be
dx dx
f (x + h) f (x)
f 0 (x) = lim
h0 h
if the limit exist. If f 0 (x) exists for each x in the interval
a < x < b, then f is said to be differentiable over a < x < b.

Summary

## There are three different interpretations of the derivative of f (x):

I Limit of the difference quotient: f 0 (x) is the limit of the
different quotient of f at x.
I Slope of the tangent line: f 0 (x) is the slope of the line
tangent to the graph of f at the point (x, f (x)).
I Instantaneous rate of change: f 0 (x) is the instantaneous rate
of change of y = f (x) with respect to x.

## MATH 1003 Calculus and Linear Algebra (Lecture 13)

Exercise

Example
Find f 0 (1) for each of the following functions:
(a) f (x) = 2x x 2
(b) f (x) = x 3
1
(c) f (x) =
x

(d) f (x) = x

Answers: a) 0 b) 3 c) -1 d) 1/2
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