Unit 1B

Limits

Lesson 4: Limits at Infinity

Determine limx→±∞5x2+12x−3.

The highest degree variable in the denominator is x. Divide all terms by x.

Because the numerator contains a radical, it is important to consider the two limits separately, one as x→∞ and the other as x→-∞.

First, find the limit as x→∞. Since x>0, x2=x=x.

limx→∞5x2+12x−3=limx→∞5x2x2+1x22xx−3x=limx→∞5+1x22−3x=5+02−0=52

Now, find the limit as x→-∞. Since x<0, x2=x=−x.

limx→−∞5x2+12x−3=limx→−∞−5x2x2+1x22xx−3x=limx→−∞−5+1x22−3x=−5+02−0=−52

Two different values suggest the lines y=52 and y=−52 will be horizontal asymptotes on the graph of the function y=5x2+12x-3, as shown below.




Note: You may have been inclined to think that two different limit values meant the limit does not exist. However, when asked to find the limits at both positive and negative infinity, you are really being asked to find two separate limits, one at positive infinity and the other at negative infinity. This is not the same as determining the limits as x approaches a value from the left and the right.
Use the graph of f(x)=2−x to predict limx→∞f(x) and limx→−∞f(x).




When x→∞, the graph of the function approaches the x-axis.

limx→∞f(x)=0.

When x→−∞, the graph does not approach the x-axis, or any particular number. Instead, as x→−∞, the value of the function increases.

limx→-∞=∞

The next Example highlights an important property of limits at infinity. The limit theorems discussed in Lesson 3 do not necessarily apply to limits at infinity because infinity is not actually a number.

Determine limx→∞x2−x.

If, incorrectly, the limit of a difference property (Property 2), introduced in Lesson 3, was used in this example involving a limit at infinity, the result would be as follows.

limx→∞x2−x=limx→∞x2−limx→∞x=∞−∞=0

However, compare this result to the graph of the function f(x)=x2−x, shown below.




As can be seen on the graph, as x→∞, the value of the function does not approach 0; in fact, the value of the function approaches infinity.

Consider why this makes sense by reviewing the initial algebraic solution.

limx→∞x2−x=limx→∞x2−limx→∞x=∞−∞=0

It can be argued that the two infinities being subtracted are not of equal magnitude since the first is really infinity squared, while the second is just infinity, suggesting the difference isn’t zero.

Instead, factor the original limit function, and then carefully make use of the limit properties.

limx→∞x2−x=limx→∞xx−1=limx→∞xlimx→∞x−1=limx→∞xlimx→∞x−limx→∞1=∞∞−1=∞

When determining limits at infinity, the method(s) and theorem(s) used must be carefully analyzed for suitability.

Consider the solution involving factoring, shown just above.

limx→∞x2−x=limx→∞xx−1=limx→∞xlimx→∞x−1=limx→∞xlimx→∞x−limx→∞1=∞∞−1=∞

In this case, a very large number is multiplied by a very large number, resulting in an even larger number, ∞.

There is an additional rule to follow when finding limits at infinity.

If n<1, then limx→∞nx=0.

Find the following limits.

a.
limx→∞34x

b.
limx→∞−23x

c.
limx→∞85x

a.
limx→∞34x

If n<1, then limx→∞nx=0. Because the n-value of 34 is less than 1, the value of the limit function will approach 0.

limx→∞34x=0

b.
limx→−∞−23x

If n<1, then limx→∞nx=0. Because −23 is less than 1, the value of the limit function will approach 0.

limx→∞−23x=0

c.
limx→∞85x

If n<1, then limx→∞nx=0. Because the n-value of 85 is greater than 1, the value of the limit function will approach ∞.

limx→∞85x=∞