MCV4U1 – UNIT ONE
UNIT ONE: LIMITS
LESSON ONE: INTRODUCTION TO LIMITS
The slope of the tangent line to a curve at a point A is the limiting slope of the secant line AB as the point
B slides along the curve towards A.
Consider the graph of y  x 2  x . Complete the table of values by letting point B approach point A.

POINT A
(1,2)
POINT B
(3,12)
(1,2)
(2,6)
(1,2)
(1.5,
(1,2)
(1.1,
(1,2)
(1.01,
(1,2)
(1.001,
SLOPE OF AB
y  y1
m 2
x 2  x1
In more general terms....
2
If A  (1,2) and B  1  h, 1  h   1  h 

Then m AB 
1  h 2  1  h   2

1 h 1
1  2h  h 2  1  h  2

h
2
3h  h

h
 3 h
1

MCV4U1 – UNIT ONE
Notice that as h gets VERY small, the slope becomes 3.
Therefore, we say the slope of the tangent at point A is....
lim slope of sec ant AB or more formally......
h0
The slope of the tangent to the graph y  f x  at a point Pa, f a is m  lim
h0
NOTE: Your textbook calls this a DIFFERENCE QUOTIENT.
Some algebra practise....
Ex. Simplify....
5  h 4  625
a)
h


b)

f a  h  f a
.
h

x  3 h  x  3
h
Ex.
Determine an expression, in simplified form, for the slope of the secant PQ .
P2,5, Q2  h, f 2  h, where f x  2x 2  3

mPQ 
f 2  h  5
2h2
22  h   3 5

h
24  4h  h 2  3 5

h
8  8h  2h 2  8

h
8h  2h 2

h
h 8  2h

h
 8  2h
2
Ex. Find the slope of the tangent to f x   x 3  2x at the point 4,72.



2
MCV4U1 – UNIT ONE
f 4  h   f 4 
m  lim
h0
h
 lim
4  h3  24  h   4 3  24 
h0
h
64  48h  12h 2  h 3  8  2h  64  8
h0
h
2
3
48h  12h  h  2h
 lim
h0
h
h 48  12h  h 2  2
 lim
h0
h
 lim 48  12h  h 2  2
 lim
h0
 50
Ex.
Find the slope of the tangent to f x  
f 1  h   f 1
h0
h
1
1

21  h  2
 lim
h0
h
1 1  h 
21  h 
 lim
h0
h
h
21  h 
 lim
h0
h
h
 lim
h0 2h 1  h 
m  lim
1
at the point when x=1.
2x

1
h0 21  h 
1

2
 lim
hw: P. 19 #4-7, 8b, 9b, 10b, 11,15, 20
3