2.1 Rates of Change
and Limits
Suppose you drive 200 miles, and it takes
you 4 hours.
Then your average speed is:
distance
x
average speed 

elapsed time
t
If you look at your speedometer during this trip, it might read 65 mph. This
is your instantaneous speed.
Vinstantaneous
y

t
200
A rock falls from a high cliff.
The position of the rock is given by:
After 2 seconds:
average speed:
y  16t
2
y  16  22  64
64 ft
ft
Vav 
 32
2 sec
sec
What is the instantaneous
speed at 2 seconds?

16  2  h   16  2 
y


t
h
2
Vinstantaneous
for some very small change
in t
2
where h = some very small
change in t

16  2  h   16  2 
y


t
h
2
Vinstantaneous
2
Look at this chart that shows “h” getting smaller and smaller…
We can see that the velocity approaches 64 ft/sec
as h becomes very small.
We say that the velocity has a limiting value of 64
as h approaches zero.
h
1
80
0.1
(Note that h never actually becomes zero.)
We can determine limits with t-charts, graphs,
and/or algebraic methods.
y
t
65.6
.01
64.16
.001
64.016
.0001
64.0016
.00001
64.0002

16  2  h   16  22
lim
h 0
h
2
The limit as h approaches
zero:
Since the 16 is
unchanged as h
approaches zero, we
can factor 16 out.
16  lim
2
4

4
h

h

4
h 0
Now let’s prove
the
instantaneous
rate
algebraically
using limits:
h
4  4h  h 2  4
16  lim
h 0
h
16  lim  4  h 
h0
0
 64

Today we will…
1.) Have an intuitive
understanding of the limiting
process
2.) Calculate limits using algebra
3.) Estimate limits from graphs
or tables of data
What is a limit????
- A limit is a statement that tells you what
height (y-value) a function is headed for as
you get close to a point specific x-value.
- It does not matter if the function actually
gets there (like if there is a hole or
asymptote at that point)
- All that matters is that you can tell where
the function INTENDS TO GO!
Sketch:
sin x
y
x
What happens as x
approaches zero?
Graphically:
Looks like y=1
So:
sin x
lim
1
x 0
x
REMEMBER THIS LIMIT, IT
COMES UP A LOT IN THE
FUTURE!!!

Basics of Limits:
Limits can be added, subtracted, multiplied, multiplied by a constant,
divided, and raised to a power.
(See your book for details.)
For a limit to exist, the function must approach the same value
from both sides.
One-sided limits approach from either the left or right side only.

The limit of a function refers to the value that the function
approaches, not the actual value (if any).
lim f  x   2
x 2
not 1

lim f  x 
x 1
2
because the left and right hand
limits do not match!
1
1
At x=1:
does not exist
2
3
4
lim f  x   0
left hand limit
lim f  x   1
right hand limit
x 1
x 1
f 1  1
value of the function

lim f  x   1
2
x 2
1
because the left and right hand
limits match.
1
At x=2:
2
3
4
lim f  x   1
left hand limit
lim f  x   1
right hand limit
x 2
x 2
f  2  2
value of the function

lim f  x   2
2
x 3
1
because the left and right hand
limits match.
1
At x=3:
2
3
4
lim f  x   2
left hand limit
lim f  x   2
right hand limit
x 3
x 3
f  3  2
value of the function

Find the following limits:
lim ( x + 4x - 3)
3
x®c
2
x + x -1
lim
2
x®c
x +5
4
2
Sketch the graph, and find the following.
ì x,
ï
f (x) = í1,
ï 2
î-x
x >1
-1 £ x £ 1
x < -1
a.)
lim f (x) = 1
x®1
b.)
c.)
lim- f (x) = -1
d.)
x®-1
lim+ f (x) = 1
x®-1
lim f (x)
x®-1
DNE since b ≠ c
How to evaluate a limit
algebraically:
1.) Simplify the function if you can
2.) Substitute the x-value into the function. If you get a
number back, that is your answer
3.) If the answer is in the form something over zero, the
answer is “undefined”
4.) If the answer is in the form 0 over 0, you have to
follow an alternative:
a.) Try Factoring to see if something cancels out
b.) Try Rationalizing if one of the terms is a square
root
c.) Try Trig Substitution
Evaluate the following limits:
Now plug in:
a.) lim x 2 - x - 2 (x - 2)(x +1)
3
x®2
4- x
2
-(x + 2)(x - 2)
Can’t plug in, we get an undefined answer, so let’s simplify first:
b.)
c.)
-
4
(x - 3)
æ
ö
x -3 =
x -3
ç
÷ (x - 3)(x + 3) ( x - 3 )
lim+ 2
x®3 x - 9
è x - 3 ø Since we still can’t plug in, THE
ì x 2 - x,
lim í
x®1 1- x,
î
LIMIT DOES NOT EXIST
x ³1
x <1
lim+ f (x) = 0
x®1
lim- f (x) = 0
x®1
Evaluate the following limits:
2
a.)
x + 2x +1 (x +1)(x +1)
lim
x®-1
b.)
x -1
2
(x +1)(x -1)
=0
(2 + x)
æ
ö
2+ x 2+ x =
ç
÷ (2 + x)(2 - x) ( 2 + x )
lim+
2
x®-2 4 - x
è 2+ x ø
Since we still can’t plug in, THE
LIMIT DOES NOT EXIST
Evaluate
a.) lim x x®2
6- x æx+
ç
x-2 èx+
x + 12 - x
lim
b.) x®-4
x+4
c.)
=
x 2 - (6 - x)
(
(x - 2) x + 6 - x
)
x2 + x - 6
6-x ö =
÷ (x - 2) x + 6 - x
(
)
6-x ø
=
(x - 2)(x + 3)
(
(x - 2) x + 6 - x
x -3
lim+
x®3
x -3
7/8
0
)
5
=
4
Evaluate the Limit:
lim
x®3
x -3
2x - 6
Since its absolute value, we
have the following:
ì x-3
,
ï
x - 3 ï 2(x - 3)
=í
2x - 6 ï -(x - 3)
,
ïî 2(x - 3)
ì1
ï 2
=í
ïî- 1
2
if x - 3 > 0
if x - 3 < 0
if x > 3
if x < 3
The limit does not exists since the two
one-sided limits are not equal
Evaluate the limit:
3- (x + Dx) - (3- x )
lim
Dx®0
Dx
2
2
3- (x 2 + 2xDx + Dx 2 ) - (3- x 2 )
lim
Dx®0
Dx
-2xDx - Dx 2
lim
Dx®0
Dx
Dx(-2x - Dx)
lim
Dx®0
Dx
= lim - 2x - Dx = -2x
Dx®0
Evaluate the following limits:
1 -1
a.)
1/25
5
x
lim
b.)
c.)
x®5
x-5
lim
x- 3
x -3
lim
x-7
x- 7
x®3
x®7
1
2 3
2 7
Limits of Trig Functions
lim cos x  1
x 0
Limits of Trig Functions
lim sin x  0
x 0
Limits of Trig Functions
lim tan x  0
x 0
Limits of Trig Functions
1
1
lim
 
x 0 sin x
0
1
lim
 

x 0  sin x
tan x
Determine lim
x®0
x
æ sin x 1 ö
= lim ç
×
÷
x®0 è x
cos x ø
sin x
1
= lim
× lim
x®0
x x®0 cos x
= 1×
1
cos0
1
= 1×
1
=1
Limits of Trig Functions
sin 2 x
lim
1
x 0
2x
Limits of Trig Functions
2 sin 2 x
sin 2 x

lim
 lim
x 0
x 0
2x
x
sin 2 x
lim 2  lim
 2 1  2
x 0
x 0
2x
Limits of Trig Functions
 x
sin  
5

lim

x 0
x
1
1
1 
5
5
1  x
sin  
5 5

x
5
Limits of Trig Functions
1
cos x
lim cot x 



x 0
sin x
0

lim   

x 0 



This video will help explain and
prove this function using the
sandwhich (squeeze) theorem
https://www.khanacademy.org/math/differentialcalculus/limits_topic/squeeze_theorem/v/proof-lim-sin-x-x
The Sandwich Theorem:
If g  x   f  x   h  x  for all x  c in some interval about c
and lim g  x   lim h  x   L, then lim f  x   L.
x c
x c
1
lim x sin    0
x 0
 x
x c
2
Show that:
The maximum value of sine is 1, so
1
x sin    x 2
x
The minimum value of sine is -1, so
1
x sin     x 2
x
So:
1
 x  x sin    x 2
 x
2
2
2
2

1
lim  x  lim x sin    lim x 2
x 0
x 0
 x  x 0
1
2
0  lim x sin    0
x 0
 x
1
2
By the sandwich theorem:
lim x sin    0
x 0
 x
2
Y=
2
WINDOW

0.02
0.01
-0.2
-0.1
0
0.1
x
0.2
-0.01
-0.02
p