Limits at Infinity Study Guide (page 3)

Updated on Oct 1, 2011

Example 7

Evaluate .

Solution 7

This has the same problem as the previous example, No matter what x may be, sin(x) will always be between –1 and 1. Thus, –1 ≤ sin(x) ≤ 1 and so

Because and , the function is squeezed between them to zero as well: . This is called the Squeeze Theorem or the Sandwich Theorem because of the way is squished between something above it going to zero and something below it going to zero.

Sign Diagrams

In order to calculate the limits at vertical asymptotes, it is necessary to know where the function is positive and negative. The key to everything is this: A continuous function cannot switch between positive and negative without being zero or undefined. Functions are generally zero when the numerator is zero and undefined where the denominator is zero. Mark these points on a number line. Between these points, the function must be entirely positive or negative. This can be found by testing any point in each interval.

For example, consider . This function is zero at x = 4 and undefined at both x= – 2 and x = 1. We mark these on a number line (see Figure 13.5).

Figure 13.5

In between x = –2 and x = 1, the function is either always positive or always negative. To find out which it is, we test a point between –2 and 1, such as 0. Because is negative, the function is always negative between –2 and 1. Similarly, we check a point between 1 and 4, such as , a point after 4, such as , and a point before –2, such as . These calculations can be made very roughly, because it matters only if the function is positive or negative at the selected point. In any case, the sign diagram for this function is shown in Figure 13.6.


Figure 13.6

This makes calculating the limits at the vertical asymptotes very easy. Not only does have vertical asymptotes at x = –2 and x = 1, but the limits are:

At the same time, we can calculate the limits at infinity:

Thus, f(x) has a horizontal asymptote of y = 0.With all of this, we begin to get a picture of , which can be seen Figure 13.7.

Figure 13.7

Notice that the horizontal asymptote y = 0 is approached from above as x → –∞, because f(x) is always positive when x < – 2. At the other end, the asymptote is approached from below as x → ∞ because the function is negative when x> 4.

Find practice problems and solutions for these concepts at Limits at Infinity Practice Questions

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