3.3 Limits and Continuity: Algebraic Approach

(This topic is also in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus)

Consider the following limit.

 limx2 x2 - 3x4x + 3 .

If you estimate the limit either numerically or graphically, you will find that

 limx2 x2 - 3x4x + 3 0.1818

But, notice that you can obtain this answer by simply substituting x = 2 in the given function:

 f(x) = x2 - 3x4x + 3 f(2) = 4 - 68 + 3 = - 211 = -0.181818...

This answer is more accurate than the one coming from numerical or graphical method; in fact, it gives the exact limit.

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, the function is continuous at the value of x in question. Recall the definition of continuity from the previous tutorial:

Continuous Functions

The function f(x) is continuous at x = a if

 limxa f(x) exists That is, the left-and right limits exist and agree with each other limxa f(x) = f(a)

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 As we saw in the previous tutorial, we can tell whether a function is continuous by looking at its graph. If the graph breaks at some point in the domain, then f has a discontinuity there. If the function is specified algrabcially, sometimes it is easy to tell whether it is continuous by just looking at the formula:

A closed-form function is any function that can be obtained by combining constants, powers of x, exponential functions, radicals, absolute values, trigonometric functions, 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)1/26x-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.

Example

 f(x) = x2 - 3x4x + 3

is a closed form function, and x = 2 is in its domain. Therefore we can obtain limx→2 f(x) by substitution:

 limx2 x2 - 3x4x + 3 = f(2) = - 211 ,

as we saw above.

 Q limx0 ex22x-3 Select one does not exist = 0 = -4/3 = -1/3 = 4/3 is undefined = -e/3 none of the above Q lim x3 2x + 3x + 3 =

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:

1. Use simplification or some other technique to replace f(x) by another closed form function which does have x = a in its domain. This allows you to substitute x = a in the new function to obtain the limit, or
2. Try evaluating the limit numerically or graphically. Note, however, that this may only give you an estimate of the limit, so we suggests trying approach (a) first.

Sometimes, you may need to use both (a) and (b).

Evaluating a Limit Using Simplification
(Similar to Example 2 in Section 3.8 of Applied Calculus, or in Section 11.8 of Finite Mathematics and Applied Calculus )

Let us evaluate

 lim x-2 3x2 + x - 10x + 2

Ask yourself the following questions:

1. Is the function f(x) a closed form function?
Answer: yes, since (3x2 + x-10)/(x+2) is a single mathematical formula.

2. Is the value x = a in the domain of f(x)?

Answer: no, since (3(-2)2+(-2)-10)/((-2)+2) is not defined.

Therefore, we consult the above Question/Answer discussion, and simplify the function, if we can.

3x2 + x - 10

x + 2
=  (x+2)(3x-5) (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:

3x - 5 lim x-2 3x2 + x-10x + 2 = lim x-2 = 3(-2)-5 = -11.

 lim x-1 (x3+1)(x-1)x2 + 3x + 2 Select one does not exist = + infinity = - infinity is undefined = 0 = -2 = -4 = -6 = -8 none of the above

Back in the tutorial for Section 1.2, we looked at the following function:

f(x)=  -1 if -4 x < -1 x if -1 x 1 x2-1 if 1 < x 2

This time, we are not showing you the graph right away, and ask you to look at the formula instead. Notice:

1. The function f is not closed form. (It is not defined by a single formula.)
2. Inside each of the separate intervals [-4, -1), (-1, 1) and (1, 2] the function is closed form, and hence continuous. Therefore, the only conceivable points where the function might fail to be continuous are on the boundaries of these intervals where we switch from one formula to another: x = -1 and x = 1.
 Q The function f is is not continuous at x = 0. Q The function f is is not continuous at x = -1. Q The function f is is not continuous at x = 1.

Now try the rest of the exercises in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus

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Last Updated: March, 2007
Copyright © 1999, 2003, 2006, 2007 Stefan Waner