# Calculus 2.1 day 1

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```					1.3 The limit of a function
A motivating example
A rock falls from a high cliff.

The position of the rock is given by:   y  16t   2

After 2 seconds:   y  16  2  642

64 ft       ft
average speed: Vav         32
2 sec      sec

What is the instantaneous speed at 2 seconds?
16  2  h   16  2 
2            2
y
Vinstantaneous       
t               h
We can use a calculator to evaluate this expression for
smaller and smaller values of h.

y
We can see that the velocity
approaches 64 ft/sec as h becomes
h       t
very small.
1           80
We say that the velocity has a limiting    0.1         65.6
value of 64 as h approaches zero.
.01        64.16
(Note that h never actually becomes         .001   64.016
zero.)
.0001 64.0016
.00001 64.0002
Definition of Limit
We write
lim f ( x)  L
x a

and say    “the limit of f(x) , as x approaches a, equals L”

if we can make the values of f(x) arbitrarily close to L
(as close to L as we like) by taking x to be sufficiently
close to a (on either side of a) but not equal to a.

In our example,                16(2  h) 2  16(2) 2
lim                          64
h 0          h
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
Left-hand and right-hand limits
We write
lim f ( x)  L
x a 
and say the left-hand limit of f(x) as x approaches a is
equal to L if we can make the values of f(x) arbitrarily
close to to L by taking x to be sufficiently close to a and
x less than a.
Similarly, we write
lim    f ( x)  L
x a 
and say the right-hand limit of f(x) as x approaches a is
equal to L if we can make the values of f(x) arbitrarily
close to to L by taking x to be sufficiently close to a and
x greater than a.
Note that
lim f ( x)  L
x a

if and only if

lim    
f ( x)  L   and   lim f ( x)  L
x a 
x a
lim f  x 
2
x 1      does not exist
because the left and
1
right hand limits do not
match!
1    2    3     4

At x=1:   lim f  x   0

left hand limit
x 1

lim f  x   1

right hand limit
x 1

f 1  1             value of the function


2                             lim f  x   1
x 2
1
because the left and
right hand limits match.
1    2     3    4

At x=2:   lim f  x   1

left hand limit
x 2

lim f  x   1

right hand limit
x 2

f  2  2            value of the function


2                            lim f  x   2
x 3
1
because the left and
right hand limits match.
1    2     3    4

At x=3:   lim f  x   2

left hand limit
x 3

lim f  x   2

right hand limit
x 3

f  3  2            value of the function



```
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