Example 1: Let f(x) = 2 x + 2 and compute f(x) as x takes values closer to 1. We first consider values of x approaching 1 from the left (x < 1).
x | f(x) |
0.5 | 3 |
0.8 | 3.6 |
0.9 | 3.8 |
0.95 | 3.9 |
0.99 | 3.98 |
0.999 | 3.998 |
0.9999 | 3.9998 |
0.99999 | 3.99998 |
We now consider x approaching 1 from the right (x > 1).
x | f(x) |
1.5 | 5 |
1.2 | 4.4 |
1.1 | 4.2 |
1.05 | 4.1 |
1.01 | 4.02 |
1.001 | 4.002 |
1.0001 | 4.0002 |
1.00001 | 4.00002 |
In both cases as x approaches 1, f(x) approaches 4. Intuitively, we say that lim
xï؟½ 1 f(x) = 4.
NOTE: We are talking about the values that f(x) takes when x gets closer to 1 and not f(1). In fact we may talk about the limit of f(x) as x approaches a even when f(a) is undefined.
Example 2: Let g(x) = sin x / x and compute g(x) as x takes values closer to 0. We consider values of x approaching 0 from the left (x < 0) and values of x approaching 0 from the right (x > 0).
x | g(x) |
-0.5 | 0.9588 |
-0.2 | 0.993346 |
-0.1 | 0.998334 |
-0.01 | 0.999983 |
-0.001 | 0.999999 |
x | g(x) |
0.5 | 0.9588 |
0.2 | 0.993346 |
0.1 | 0.998334 |
0.01 | 0.999983 |
0.001 | 0.999999 |
Here we say that lim
xï؟½ 0 g(x) = 1. Note that g(0) is undefined.
Example 3: The graph below shows that as x approaches 1 from the left, y = f(x) approaches 2 and this can be written as
lim
xï؟½ 1- f(x) = 2
As x approaches 1 from the right, y = f(x) approaches 4 and this can be written as
lim
xï؟½ 1+ f(x) = 4
Note that the left and right hand limits and f(1) = 3 are all different.
Example 4: This graph shows that
lim
xï؟½ 1- f(x) = 2
As x approaches 1 from the right, y = f(x) approaches 4 and this can be written as
lim
xï؟½ 1+ f(x) = 4
Note that the left hand limit and f(1) = 2 are equal.