Lecture 3
Intuitive Limits
Now We Start Calculus
The Problem of Tangents
Tangents to a Circle
General line through P
P( xo, yo )
Radius
Tangent Line
C = (h,k)
yo  k
Slope of radius =
xo  h
xo  k
Slope of tangent =  y   h
o
Finding the Tangent Line at a Point
P( xo, yo ) to a circle of radius r,
centered at C = (h,k) using algebra
• Point slope form:
• Suffices to find the
slope
• Key – tangent meets
the circle only in at the
point P.
yyom ( xxo )
Find the points of Intersection
• Circle:
• Line
2
2
2
( xh ) ( yk ) r
yyom ( xxo )
• Substitute for y
2
2
2
( x h )  ( yo m ( x xo ) k )  r
Illustrate what Happens for (h,k) =
(0,0)
2
2
2
x  ( yo m ( x xo ) )  r  0
2
2
2
2
2
( m  1 ) x  (  2 m xo 2 yo m ) x yo 2 m xo 2 2 yo m xo r  0
LHS must equal
2
( m  1 ) ( x xo ) 2
Subtract and get
2
2
( 2 yo m 2 xo ) x yo  2 yo m xo r  xo 2 0
So
2 yo m 2 xo 0
m   
xo
yo
Slope of Secant
• Connects (a, f(a)) with (a+h, f(a+h)) so slope is
f( a h ) f( a )
(a+h) a
f( a h ) f( a )
h
Go through Similar Process With
Some Functions
f( x )x
2
Geometric Idea:
Tangent at P meets the
tangent line at only at P
(near P)
Tangent to graph of
at the point ( xo, f( xo ) )
f( x )x
yf( xo )m ( xxo )
2
x  xo 2 m ( x xo )
(  m2 xo ) xm xo2 xo 20
m  2 xo
2
Problem:
Such calculations are very difficult for
more complicated functions –
impossible for others. Need a new idea
• Don’t vary the slope of the general line – vary
the other point
• The “secant line” should approximate the
tangent line
• If one line “approximates” another then it’s
slope should approximate that of the other line.
Tangent Line at (a,f(a))
Approximating Secant Line
Slope of the secant line is
f( a h ) f( a )
h
Idea is that as Q gets closer and closer to P (i.e. as h
gets closer and closer to 0) the slopes of the secant
lines get closer to that of the tangent line
We denote this
• Slope of the tangent line to graph of f at
x = a is
lim
h  0
f( a h ) f( a )
h
Notation:
• If f is a function and a is in its domain then the
slope of the tangent line to the graph of f(x) at the
point (a,f(a)) is denoted f ‘ (a) so we write
f ‘ (a)
=
lim
h  0
f( a h ) f( a )
h
Calculate f ‘(4) if
f ‘ (a) =
lim
h  0
Here a = 4
f ‘ (4) =
f ‘ (4) =
lim
h  0
lim
h 0
f( x )x
f( a h ) f( a )
h
f( 4 h ) f( 4 )
h
2
2
( 4 h )  4
h
2
lim
h 0
2
16 8 h h  16
lim
h 0
h
8 h h
2
h
lim 8 h
h  0
= 8
Calculate f ‘(a) for any a
f ‘ (a) =
=
=
=
lim
h  0
lim
h 0
lim
h 0
f( a h ) f( a )
h
2
2
( a h )  a
h
2
2
2
a  2 a h h  a
h
lim 2 a h 2 a
h  0
=
lim
h 0
2 a h h
h
2
To do these must be able to
calculate expressions of form
lim f( x )
xa
Taken to (intuitively mean) :
The value to which f(x) tends as x gets closer
and closer to ( but never equals) a
Facts about Limits
• Limits may or may not exist
– this limit does not exist
• There is an algebra of
limits provided they exist
lim
x0
1
x
Some Basic Limits
• If c is a number and a
is any number then
lim c c
x  a
n
n
x  a
• If n is a number and a
>0 then
lim
x a
• If f(x) and g(x) agree
except at x = a then
lim f( x )  lim g( x )
x  a
x  a
If
lim f( x )
xa
and
lim g( x )
xa
both exist
then
lim f( x ) g( x ) ( lim f( x ) ) ( lim g( x ) )
x  a
x  a
x  a
and
lim f( x ) g( x ) ( lim f( x ) ) ( lim g( x ) )
x  a
x  a
x  a
The basic limit theorems
• In general calculating a limit “from scratch”
is difficult.
• The limit theorems allow us to calculate
new limits from old without having to
repeat what has already been done.
If
lim f( x )
xa
exists and c is a number then
lim c f( x ) c ( lim f( x ) )
x  a
x  a
lim f( x ) and
xa
If
and
lim g( x )
xa
f( x )
lim g( x ) both exist
xa
is not zero then
lim f( x )
x a
lim
 
g( x )
lim g( x )
x a
x a