Calculus 2.1 day 1

Document Sample
Calculus 2.1 day 1 Powered By Docstoc
					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


                                                             

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:0
posted:11/4/2012
language:Unknown
pages:10