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Consider the following limit.
x 2 |
 4x + 3 |
. |
If you calculate the limit either numerically or graphically, you will find that
| x2 |
4x + 3 | = - 0.181818... = - |  11 | . |
But, notice that you can obtain the same asnwer by simply substituting x = 2 in the given function:
| f(x) = |
4x + 3 | ||||
| f(2) = |
 8 + 3 | = | - | 11 | . |
Q Is that all there is to evaluating limits algebraically: just substitute the number x is approaching in the given expression?
A Not always, but this often does happen, and when it does, we say that the function is continuous at the value of x in question.
|
Continuous Functions
The function f(x) is continuous at x = a if
The function f is said to be continuous on its domain if it is continuous at each point in its domain. If f is not continuous at a particular a, we say that f is discontinuous at a or that f has a discontinuity at a. |
Q How do you tell if a function is continuous?
A Somewtimes, it is easy: A closed-form function is any function that can be obtained by combining constants, powers of x, exponential functions, radicals, logarithms (and some other functions you may never see) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. Examples of closed-form functions are:
| 3x2 - x +1, |
| (x2 - 1)1/2 6x-2 |
| e(x2 - 1)1/2/x |
| (log3(4x2 - ex))2/3 |
They can be as complicated as you like. The following is not a closed form function.
| f(x) = |  |
-1 if x < -1 |
x2 + x if -1  x 1 |
||
| 2 - x if 1 < x 2 |
The reason for this is that f(x) is not specified by a single mathematical expression. What is nice about closed-form functions is the following.
|
Continuity of Closed Form Functions
Every closed form function is continuous on its domain. Thus, the limit of a closed-form function at a point on its domain can be obtained by substitution. Quick Example
is a closed form function, and x = 2 is in its domain. Therefore we can obtain x
as we saw above. |
Q What if f(x) is a closed form function, but the point x = a is not in the domain of the function?
A Then, you either:
Sometimes, you may need to use both (a) and (b).
Evaluating a Limit Using Simplification
(Similar to Example 2 in Section 3.7 of Applied Calculus, or in Section 11.7 of Finite Mathematics and Applied Calculus )
Let us evaluate
x 2 |
 x + 2 |
Ask yourself the following questions:
1. Is the function f(x) a closed form function?
2. Is the value x = a in the domain of f(x)?
Therefore, we consult the above Question/Answer discussion, and simplify the function, if we can.
x + 2 | = |
| |
| = | 3x-5. |
Since we are now left with a closed form function that is defined when x = -2, we can now evaluate the limit by substitution:
| x2 |
x + 2 |
= |
x2 |
3x - 5 | = | 3(-2)-5 = -11. |
Now try some of the exercises in Section 3.7 of Applied Calculus or Section 11.7 of Finite Mathematics and Applied Calculus.