Maths Quest Maths B Year 11 for Queensland
Chapter 13 Differentiation and applications WorkSHEET 13.1
WorkSHEET 13.1 Differentiation and applications
1
Evaluate the following limits:
(a) lim ( x  2)
(a)
lim ( x2  4 x  4)
(c)
lim ( x  8)
x 2
Name: _____________________
lim ( x  2)
1
x2
as x  2, ( x  2)  2  2  0
x2
(b)
1
(b)
3
lim ( x2  4 x  4)
1
x 2
as x  2,
x 2
( x 2  4 x  4)  2 2  4  2  4  0
(c)
lim ( x3  8)
1
x 2
as x  2, ( x 3  8)  2 3  8  0
2
Evaluate the following limits:
 x2  4 

(a) lim 
x 2 x  2 


(b)
(a)
 x3  8 

lim 
x2 x  2 


 x2  4 

lim 
x2
 x2 
 ( x  2)( x  2) 
 lim 

x2
x2

 lim  x  2 , x  2
1
1
x2
4
(b)
 x3  8 

lim 
x2
 x2 
 ( x  2)( x 2  2 x  4) 
 lim 

x2
x2


1
 lim x 2  2 x  4 ,
1
x2
 12


x2
Maths Quest Maths B Year 11 for Queensland
3
Evaluate the following limit:


x3  64
lim  2

2
x  2  x  16 x  4 x  16 



Chapter 13 Differentiation and applications WorkSHEET 13.1
x 3  64
x 2  16 x 2  4 x  16


2


 x  4 x  4 x  16
 x  4 x  4 x 2  4 x  16

1
x4
2
x  4,4
1
So


x 3  64
lim  2

2
x  2  x  16 x  4 x  16 
1
 lim
x2 x  4
1

6

4
For what values of x is the function in
(a) question 2 (a) discontinuous?
(b)
question 2 (b) discontinuous?
(c)
question 3 discontinuous?
(a)
(b)
(c)
5
Investigate the following limit:
x  [0, 2]
 2  x,
lim f ( x)  
2
x 2
x  (2, 6]
4  x ,
6
Investigate the following limit:
x  [3, 6]
 2 x,
lim f ( x)   2
x 3
x  [3, 3)
 x  9,


 x2  4 

 is discontinuous when the
x

2


denominator is undefined; that is,
when x = 2
 x3 8 
 is discontinuous
Similarly, 
 x2 
when x = 2.


x3  64
 2

2
 x  16 x  4 x  16 
is discontinuous when
x 2  16  0
x  4



1
1
1
1
As x  2, 2  x  0
1
As x  2, 4  x2 0
As both parts of the function approach the same
limit, the limit exists.
For lim f  x   2 x  2  3  6
x 3
For lim f x   x2  9  32  9  0
x 3
As both parts of function do not approach the
same limit, the limit does not exist.
1
Maths Quest Maths B Year 11 for Queensland
7
Chapter 13 Differentiation and applications WorkSHEET 13.1
Find f x  h for each of the following:
(a)
f x   3x  6
(b)
(a)
f x   x2  4 x  3
f  x   3x  6
f  x  h   3x  h   6
 3x  3h  6
f x   x 2  4 x  3
(b)
f  x  h    x  h   4 x  h   3
2
3
1
1
 x 2  2 xh  h 2  4 x  4h  3
8
(a)
What is the gradient of the straight line
with equation: f x   4 x  7 ?
(b)
Compare the answer to part (a) with the
answer to the following limit:
f x  h   f x 
lim
h 0
h
What can you conclude?
(a)
f x   4 x  7 The gradient is equal to
the coefficient of x, which is  4.
(b)
lim
f x  h   f x 
h
 4 x  h   7   4 x  7 
 lim
h 0
h
 4 x  4h  7  4 x  7
 lim
h 0
h
 4h
 lim
h 0
h
 lim (4)
h0
h 0
1
1
h 0
 4
The same answer indicates that the expression
in part (b) is an alternative way of deriving the
gradient.
1
Maths Quest Maths B Year 11 for Queensland
9
Chapter 13 Differentiation and applications WorkSHEET 13.1
f x  h   f x 
to find an
h
expression for the gradient for each of the
following:
Use the limit lim
(a)
h 0
(a)
f x   7 x  3
(b)
f x   x2  2 x  8
4
2
f x  h  f x 
lim
h 0
h
7 x  h   3  7 x  3
 lim
h0
h
7 x  7h  3  7 x  3
 lim
h0
h
7h
 lim
h0 h
 lim 7
h0
h0
7
(b)
f x  h  f x 
h
 x  h 2  2x  h   8  x 2  2 x  8
 lim
h 0
h
2
2
x  2 xh  h  2 x  2h  8  x 2  2 x  8
 lim
h 0
h
2
2 xh  h  2h
 lim
h 0
h
h2 x  h  2
 lim
h 0
h
 lim 2 x  h  2
h0
lim
h 0
h 0
 2x  2


2
Maths Quest Maths B Year 11 for Queensland
10
Chapter 13 Differentiation and applications WorkSHEET 13.1
Evaluate f  3 for each of the following:
(a)
f  x    x 2  3x
(b)
f x   x  x
3
(a)
2
f x  h  f x 
f  x   lim
h 0
h
2
  x  h   3 x  h    x 2  3x
f   x   lim
h 0
h
2
2
 x  2 xh  h  3x  3h  x 2  3x
 lim
h 0
h
2
 2 xh  h  3h
 lim
h 0
h
h 2 x  h  3
 lim
h 0
h
 lim (2 x  h  3)
h0

2
5

h 0
 2 x  3
So f  3  2  3  3  3
(b)
f  x   lim
x  h
2
h 0
 lim
3

 x  h  x  x
h
3
2
2
3
3
2
2

2
2
3
x  3 x h  3 xh  h  x  2 xh  h  x  x
h 0
h
3 x h  3xh  h  2 xh  h 2
 lim
h 0
h
2
h 3 x  3 xh  h 2  2 x  h
 lim
h 0
h
2
 lim (3x  3 xh  h 2  2 x  h) h  0
2
2
3

h 0
 3x 2  2 x
So f  3  3  (3)2  2  3  33

2