Math Glossary
Limit
Definition: Limit from the Left
The limit as x approaches c from the left of f(x), denoted
| Lim | f(x) |
| x→c- |
is the value that f(x) approaches when x gets closer and closer to c from the left without getting to c. In other words, keeping x<c, move x toward c. To what number does f(x) get arbitrarily close?
Definition: Limit from the Right
The limit as x approaches c from the right of f(x), denoted
| Lim | f(x) |
| x→c+ |
is the value that f(x) approaches when x gets closer and closer to c from the right without getting to c. In other words, keeping x>c, move x toward c. To what number does f(x) get arbitrarily close?
Definition #1: Overall Limit
The limit as x approaches c of f(x), denoted
| Lim | f(x) |
| x→c |
is the value that f(x) approaches when x gets closer and closer to c from both left and right without getting to c. In other words, keeping x≠c, move x toward c from either side. To what number does f(x) get arbitrarily close?
Definition #2: Overall Limit
If
| Lim | f(x)= | Lim | f(x)=L |
| x→c- | x→c+ |
then
| Lim | f(x)=L |
| x→c |
Definition #3: Overall Limit ( δ-ε definition)
| Lim |
f(x)=L |
| x→c |
where L is a real number, if and only if for any ε>0 there exists a δ>0 such that when 0<|x-c|<δ, then |f(x)-L|<ε.
Examples:
#1
| For f(x)= | { | x2 | ,x<1 |
| 7 | ,x=1 | ||
| x+3 | ,x>1 |
The graph from the left of x=1 approaches the point (1,1). So
| Lim | f(x)= | Lim | x2=12=1 |
| x→1- | x→1- |
The graph from the right of x=1 approaches the point (1,4). So
| Lim | f(x)= | Lim | (x+3)=1+3=4 |
| x→1+ | x→1+ |
The graph from the left and of x=1 approaches 2 different points. So
| Lim | f(x) |
doesn't exist. |
| x→1 |
Note, nowhere was I concerned about the point (1,7) on the graph. Limits are only concerned with what is happening to the left and/or right of x=c, not with what is happening at x=c.
#2
| For f(x)= | { | 5 | ,x=0 |
| 2x+1 | ,x≠0 |
The graph from the left of x=0 approaches the point (0,1). So
| Lim | f(x)= | Lim | (2x+1)=2(0)+1=1 |
| x→0- | x→0- |
The graph from the right of x=0 approaches the point (0,1). So
| Lim | f(x)= | Lim | (2x+1)=2(0)+1=1 |
| x→0+ | x→0+ |
The graph from the left and right of x=0 approaches the point (0,1). So
| Lim | f(x)=1 |
| x→0 |
Note that the limits are 1 while f(0)=5.
#3 Prove that
| Lim | (x2-3)=1 |
| x→2 |
Thus, set f(x)=x2-3, c=2 and L=1. Assume that 0<|x-2|<δ. Then,
| |f(x)-L| | = | |(x2-3)-1| |
| = | |x2-4| | |
| = | |x+2| |x-2| | |
| < | |x+2| δ |
Since we think of δ and ε as being small numbers, we can also assume that δ<1. This makes
|x-2|<δ<1
|x-2|<1
-1<x-2<1
1<x<3
3<x+2<5
Hence
| |f(x)-L| | < | |x+2| δ |
| < | 5δ |
Now if we set 5δ=ε and solve for δ giving δ=ε/5.
| |f(x)-L| | < | 5δ |
| < | 5(ε/5) | |
| < | ε |
Therefore, for any given ε>0, we can set δ equal to the smaller of 1 and ε/5. Then whenever 0<|x-2|<δ, |f(x)-L|<ε.
See also:
EMAIL:rickmanw@seminolestate.edu
Seminole State Math Student Website