BC 1
Limits 6
Name:
When we write lim   x   L , we mean that ƒ(x) gets close to L when x gets close to a.
x a
But how close? Very close. Oh... What's "very close" mean? Well, if L really is the limit, then
the value of (x) must be arbitrarily close to L when x gets close enough to a.
Definition: We say that the limit of (x) as x approaches a is L if and only if for each   0 ,
there exists a   0 such that 0  x  a   implies 0  f ( x)  f (a)   .
Notes: i. Here (epsilon) is a letter in the Greek alphabet used to represent some (arbitrarily)
small positive number and  (delta) is also a letter in the Greek alphabet.
ii. This definitions states that if L is the limit, then we can force f ( x) to be as close to L as
we wish (no further than by requiring x to be sufficiently close to a (within  )
(1)
Let (x) = 2x + 1 with a = 1, so that lim  2 x  1  3 .
x1
(a)
If  = 0.5.
(x) = 2x + 1
Find a value of  such that if x  (1   ,1   ) ,
then f ( x)  (3   ,3   )  (2.5,3.5) .
L +  = 3.5
L=3
L –  = 2.5
a
Note that this doesn’t prove lim  2 x  1  3 , We must show that we can find such a
x1
number   0 for any positive  , no matter how small.
Proof: We must show that for each   0 , there exists a   0 such that 0  x  1  
 2 x  1  3   .
To that end, let   0 be an arbitrary number. Choose  
0  x 1   
 2 x  1  3 
2x  2
 2 x 1
 2

Therefore, we have shown lim  2 x  1  3
x1

2
. Then
implies
BC 1
Limits 6

Name:

Example: We claim lim x 2  x  2 .
x2
Proof: We must show that for each   0 , there exists a   0 such that 0  x  2  
x
2
 x  2   .
To show this, we let let   0 be an arbitrary number.

Choose   minimum of the numbers 1 or . (Where does this come from)?
3
Then we first note that if   1, it follows that x  2  1  1  x  3 . So,
0 x2  
x
2
 x   2  ( x  2)( x  1)

 x  2 x 1
   3


Therefore, we have shown lim x 2  x  2
x 2
Exercises:
6
1. Prove: lim    3 .
x 2  x 
implies