MAT 3749 2.1 Part 1 Handout
Left-Hand Limit
We write lim f ( x)  L and say “the left-hand limit of f ( x) , as x approaches a, equals
xa
L” if we can make the values of f ( x) arbitrarily close to L (as close as we like) by taking
x to be sufficiently close to a and x less than a.
Right-Hand Limit
We write lim f ( x)  L and say “the right-hand limit of f ( x) , as x approaches a, equals
x a
L” if we can make the values of f ( x) arbitrarily close to L (as close as we like) by taking
x to be sufficiently close to a and x greater than a.
Limit of a Function
lim f ( x)  L if and only if lim f ( x)  L and lim f ( x)  L
x a
xa
x a
1
Recall lim sin   DNE
x 0
x
Deleted Neighborhood*
For all practical purposes, we are going to use the following definition:
Let I be an open interval with a  I . I  a is called a deleted neighborhood of a .
Last Edit: 2/16/2016 3:15 AM
Precise Definition
We say that f  x  approaches the limit L as x approaches a and write
lim f  x   L ,
x a
if f is defined on some deleted neighborhood of a and,   0 ,   0 such that
f  x  L  
if
0 xa  .
L
L
L
L
L
a
a
aa
a


The values of  depend on the values of  .  is a function of  .
Definition is independent of the function value at a .
2
Example 1
Use the    definition to prove that lim  2 x  3  5 .
x 1
Analysis
Proof
Guessing and Proofing
Note that in the solution of Example 1 there were two stages - guessing and proving.
We made a preliminary analysis that enabled us to guess a value for  . But then in the
second stage we had to go back and prove in a careful, logical fashion that we had made a
correct guess. This procedure is typical of much of mathematics. Sometimes it is
necessary to first make an intelligent guess about the answer to a problem and then prove
that the guess is correct.
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Example 2
Use the    definition to prove that lim x 2  9 .
x 3
Analysis
Proof
4
Example 3
1
Prove that lim sin   DNE.
x 0
x
Analysis
5
Classwork
The following steps help you to provide a proof for example 3.
(a) Solve sin x  1 for x  0 . Write your answers in terms of n N0  0  N .
1
(b) Find a sequence of number an  such that sin    1 and an are closed to zero
 an 
when n is sufficiently large.
an 
for n  N0 .
(c) (a) Solve sin x  1 for x  0 . Write your answers in terms of n  N0 .
1
(d) Find a sequence of number bn  such that sin    1 and bn are closed to zero
 bn 
when n is sufficiently large.
bn 
for n  N0 .
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(e)
Proof
1
1
Suppose lim sin   exists and lim sin    L .
x

0
x 0
x
x
1
1
1
. Then   0 such that sin    L  whenever 0  x  0   , i.e.
2
2
x
0 x  .
Let  
Let an 
and bn 
for n  N0 .
For sufficiently large n , an   and bn   .
 1
So, sin 
 an

L 

1
and sin 
 bn

L 

.
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And
 1
sin 
 an

1
  sin   

 bn 
 Use triangle
inequality to show that
 1 
1
sin    sin    1
 an 
 bn 
Be sure to provide
details.
1
However,
 1
1
sin    sin   
 an 
 bn 
Show that
 1 
1
sin    sin    2
 an 
 bn 
Be sure to provide
details.
2
which is a contradiction to the previous inequality.
1
Therefore, lim sin   DNE.
x 0
x
Specify the exact
contradiction.
State the conclusion.
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