Download limits and derivatives conditions for the existence of a limit

LIMITS AND DERIVATIVES
The limit of a function is defined as the value of y that the curve approaches, as x
approaches a particular value.
The limit of f (x) as x approaches a is written as Lim f ( x) , and represents the value that
x a
f ( x) approaches, as x gets closer and closer to a .
CONDITIONS FOR THE EXISTENCE OF A LIMIT
A limit will exist at x  a if:


If the function f  x  is continuous at x  a .
If there is point discontinuity at x  a .
A limit will not exist at x  a if:

There is discontinuity across an interval and x  a lies within this interval.
A limit does not exist
at this value of x
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A limit does not exist
at this value of x
The Essentials – Year 11 Mathematical Methods – Book 1
Page 1
EVALUATING LIMITS GRAPHICALLY
SINGLE FUNCTIONS
Step 1: Observe the value of y as x approaches a value below the point of interest.
Step 2: Observe the value of y as x approaches a value above the point of interest.
Step 3: If the left hand limit is equal to the right hand limit a limit exists at x  a .
The limit is simply equal to the value of y that the curve is approaching on either
side of the point of interest.
QUESTION 1
The graph of f ( x) is given on side.
y
lim f  x  is equal to
x 1.5
A
B
C
D
E
0
3
3
1.5
x
Undefined
Solution
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 2
LIMITS OF HYBRID FUNCTIONS
A hybrid function is a function that has different rules describing the different sections of its
domain.
Limits of hybrid functions are evaluated in the same manner as previously described.
Step 1: Find the limit of each individual function at the given value of x .
Step 2: If the limiting values are the same, the limit exists at that value of x .
If one or more of the limiting values are different, the limit does not exist at that
particular value of x .
If the function is discontinuous across an interval and the value of x lies in this
interval, the limit cannot be evaluated
Note:
QUESTION 2
The graph of f (x) is shown below.
(a)
Find lim f ( x)
(b)
Find lim f ( x)
(c)
lim f ( x)  lim f ( x)
x 5
x 2
x 1
x 3.5
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 3
EVALUATING LIMITS ALGEBRAICALLY
Does the given equation contain a denominator bearing terms involving x ?
No
Yes
Substitute the limit value into
the given equation and solve.
Substitute the limit value into
the denominator.
If the denominator is equal to zero, factorise
and eliminate terms by cancellation.
Then substitute the limit value into the
simplified equation and solve.
If the denominator is not equal to
zero, substitute the limit value into
the given equation and solve.
The manner in which we evaluate limits algebraically depends on whether the denominator
of the function carries terms involving x :
Given Lim f (x) :
xa
(a)
If the equation does not carry a denominator, or if the denominator has NO terms
involving x , simply substitute the value of a into the given equation and simplify.
For example: lim(5  2 x  x 3 )  5  2(1)  (1)3  2
x 1
(b)
If the denominator carries terms involving x , we need to first determine whether the
value of x (or a ) in question causes the function to be undefined.
Substitute the value of a into the denominator of the equation.
If the answer does not equal zero, proceed to substitute the value of a into the
equation.
 x 2  1
( 2) 2  1
4 1
5
 
For example: lim 

 , x2
x  2
4
4
 x  2  (2)  2
If the answer is equal to zero, we need to eliminate the term(s) that is/are causing the
function to become undefined. Factorise the given equation and eliminate these terms
by cancellation. Before evaluating the limit, substitute the value of a into the new
equation to ensure there are no other terms present that will make the function
undefined.
 ( x 2  5 x  6) 
 ( x  3)( x  2) 
For example: lim 
  lim 
  lim ( x  2)  3  2  1
x 3
 ( x  3)  x 3  ( x  3)  x 3
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 4
Note:
When there are terms involving x in the denominator of a fraction, there may be some
values of x which cause the function to become undefined. These values of x must be
stated with your answer.
To determine which values of x will make the function undefined, before the function is
factorised and terms are eliminated by cancellation:
Step 1: Let the bottom of fraction equal zero .
Step 2: Solve for x .
 x 2  3x  4 
 Restrictions: x  1 or x  3 .
For Example: Lim 
x  1 ( x  1)( x  3) 


Note: Even though this function is undefined at x  1 , we may still investigate what
happens to the value of y as x approaches this value (i.e. the limit).
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 5
LIMIT THEOREMS
(a)
The limit of a constant is equal to the value of the constant:
If f ( x)  k then Lim f ( x)  k .
xa
For Example: Lim 5  5
x 2
(b)
The limit of the sum and/or difference of a series of terms is equal to the sum
and/or difference of the limits of each individual term.
Lim f  x   g  x   Lim f  x   Lim g  x 
x a
x a
xa


 
For Example: Lim x 2  3 x  1  Lim x 2  Lim3 x   Lim1  4  6  1  3
x2
(c)
x2
x2
x2
The limit of the product of two functions is equal to the product of the
limits of each individual function.
Lim f  x .g  x   Lim f  x  Lim g  x 
xa
For Example:
(d)
xa
xa




4 x   0  4  0
Lim
x 2  1 4 x   Lim
x 2  1  Lim



x 1
x 1
x 1
The limit of the quotient of two functions is equal to the quotient of the limits of
each individual function.
 f  x   Lim f  x 
Lim 

 g  x   Lim g  x 
Note that g  x   0 .
Lim x  5
x5
2 1
2


x 3





2
x 3 2 x 2  9 x  5
2 x  9 x  5 18  27  5  14 7
 Lim

x 3
For Example: Lim
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

The Essentials – Year 11 Mathematical Methods – Book 1
Page 6
QUESTION 3

x 2  25 
 , if it exists.
2
 x  2 x  15 
Evaluate lim 
x 5
Solution
Substitute x = -5 into denominator to determine if the function is defined at this point:
Denominator = 0
Factorise and eliminate terms by cancellation:
 x 2  25 
( x  5)( x  5)
( x  5)
 lim
 lim
lim  2

x 5 x  2 x  15

 x 5 ( x  3)( x  5) x5 ( x  3)
Evaluate the required limit - Substitute x = 3 into the simplified expression:
( x  5) 5

x 5 ( x  3)
4
lim
State any restrictions on the values of x:
x 3  0
x  3
QUESTION 4
Evaluate lim 14  6t  t 3 , if it exists.
t 2


Solution
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 7
QUESTION 5
 x2  1
 .
Evaluate lim 
x  2
 x2 
Solution
QUESTION 6
 (3  h) 2  9 
lim 
 is equal to:
h 0
h


A
B
C
D
E
0
undefined
6h  h 2
6
h
QUESTION 7
Show that lim
z 0
z 2  6  lim z 2  lim 6 .
z 0
z 0
Solution
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 8
QUESTION 8
Given lim f ( x)  9 , lim g ( x)  2 and lim h( x)  4 , determine the following limits.
x 8
x 8
(a)
lim  2 f ( x)  12h( x) 
(b)
lim  g ( x).h( x)  f ( x) 
(c)
lim  f ( x)  g ( x)  h( x)
x 8
x 8
x 8
x 8
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 9
DIFFERENTIATION
The derivative describes the gradient of the tangent to a curve at any value of x .
The process of finding a derivative is called differentiation.
Tangent
The gradient of a line is constant and does not change at different values of x . As the
derivative represents the gradient to a curve, the derivative of a linear expressions will result
in a constant (numeric) value.
The gradient to a curve modelled by a non linear equation changes, and is dependent on
the value of x .
Therefore, the derivative of a non linear expression will result in an expression in terms
of x . Only when a particular value of x is substituted into the derivative equation, will a
numeric value be obtained. This value represents the gradient of the tangent to the curve at
the specific value of x .
When the equation is denoted as f (x ) , the derivative is represented as f ' ( x ) .
When the equation is denoted as y , the derivative is represented as
dy
.
dx
Note:
dy
is an operation - it is not a quotient (it is not the same as dy  dx).
dx
Note:
d

dx
 reads “the derivative of   with respect to
x.
Derivatives may be evaluated from first principals or by using a set of rules.
Whenever you see the following words/phrase: gradient, gradient function,
gradient of the tangent, automatically think of differentiation.
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 10
DERIVATIVES FROM FIRST PRINCIPLES
We can use limits to find derivatives using a process that is referred to as “differentiation
using first principles”. The derivative from first principles is obtained by applying the rule:
Derivative  Lim
h 0
[ f ( x  h)  f ( x)]
dy d
d
 f '( x) 
 ( y )   f ( x) 
h
dx dx
dx
An alternative notation for the limit theorem is:
[ f ( x   x)  f ( x)]
δx 0
x
lim
where x is a small increment (change) in x .
FINDING THE DERIVATIVE FROM FIRST PRINCIPLES
METHOD:
Step 1: Write the expressions for f (x) and f ( x  h) .
Step 2: Substitute the expressions for f (x) and f ( x  h) into the limit theorem.
[ f ( x  h)  f ( x)]
dy
 Lim
h

0
dx
h
Step 3: Expand and collect like terms.
Step 4: Remove h as a common factor and simplify.
Step 5: Substitute h  0 .
Note:


f (x) represents an equation in terms of x .
To obtain f ( x  h) from f (x) , replace x in the given equation with ( x  h) .
For Example: If f ( x)  x 2  6 x  1 , then f ( x  h)  ( x  h) 2  6( x  h)  1 .

To find the derivative at a specific value of x , substitute the given value of x into the
derived equation.
Note: The gradient at x  2 is denoted as f ' (2) .
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 11
QUESTION 9
Find the derivative of f ( x)  x 2  3x using first principles. Hence find the gradient of the
tangent to the curve at x  2 .
Solution
Write the expressions for f (x) and f ( x  h) :
If f ( x)  x 2  3x then f ( x  h)  ( x  h) 2  3( x  h)
Substitute the expressions for f (x) and f ( x  h) into the limit theorem:
 f ( x  h)  f ( x ) 
f ( x)  lim 

h0
h

 ( x  h) 2  3( x  h)  ( x 2  3 x) 
 lim 

h 0
h


Expand and collect like terms:
 x 2  2 xh  h 2  3 x  3h  x 2  3 x 
 lim 

h 0
h


 2 xh  h 2  3h 
 h(2 x  h  3) 
 lim 
 lim 
x  h  3)  2 x  3,

  lim(2
h 0
h 0
h
h

 h 0 
h0
The derivative or the equation describing the gradient to the curve y  x 2  3 x is 2 x  3 .
Substitute in the value of x :
When x  2 , the gradient is equal to f (2)  1 .
NOTE:

The lim notation was included in each step, until h  0 was substituted into the equation.

You may check your answers by differentiating the given equation by rule. If the two
answers are different, it is likely that an error has been made when expanding the
brackets in the limit formula. Remember to multiply every term in the second set of
brackets by negative one!
h0
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 12
QUESTION 10
(a)
Show, using first principles, that the derivative of f ( x) 
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x
1
is f '( x) 
.
( x  1) 2
x 1
The Essentials – Year 11 Mathematical Methods – Book 1
Page 13
(b)
Hence find f '(5) .
(c)
Find the coordinates of the point(s) on the curve at which the gradient of the tangent
equals 4 .
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The Essentials – Year 11 Mathematical Methods – Book 1
Page 14