Limits

Example 1: Finding a Rectangle of Maximum Area

You are given 24 inches of wire and are asked to form a rectangle 

whose area is as large as possible. What dimensions should the rectangle have?

Let w represent the width of the rectangle and let l represent the length of the rectangle. 

Because 2w + 2l = 24 (perimeter is 24), it follows that l = 12-w and that the area of the rectangle is 

A = wl --> Formula for area

A = w(12 - w) --> Substitute 12 - w for l

A = 12w - w^2 --> Simplify

Using this model for area, you can experiment with difficult values of w to see how to obtain the maximum area. After trying several values, it appears that the maximum area occurs when w = 6.

Example 2 : Finding a Limit That Can Be Reached

Use a table to estimate numerically the limit lim x--> 2 (3x - 2)

Let f(x) = 3x - 2. Then construct a table that shows values of f(x) when x is close to 2.

From the table, it appears that the closer x gets to 2, the closer f(x) gets to 4. Thus, you can estimate the limit to be 4. For this particular function, you can obtain the limit simply by substituting 2 for x to then obtain lim x--> 2 (3x - 2) = 3(2) - 2 = 4. 

Conditions Under Which Limits Do Not Exist:

The limit of f(x) as x--> c does not exist if any of the following conditions is true.

1. f(x) approaches a different a different number from the right side of c than from the left side of c.

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

Example 4: Oscillating Behavior

Discuss the existence of the limit.

lim x--> 0 sin (1/x) 

Let f(x) = sin(1/x). In the diagram above, you can see that as x approaches 0, f(x) oscillates between -1 and 1. Therefore, the limit does not exist because no matter how close you are to 0, it is possible to choose values of x(1) and x(2) such that sin (1/x(1)) = 1 and sin (1/x(2)) = -1, as indicated in the table.