Calculus I, Chapter 3: The Average Rate of Change and Limits
Previous Section: 3.9 The Squeeze Theorem
At this point, we have discussed each of the major types of limits that occur and how to navigate them. The last question that arises is how to deal with limits at infinity. These limits are really asking what happens at the tail ends of a function, as the x-value gets very large in either the positive or negative direction. This can be helpful for identifying trends over a large period time and specifically pinpointing a value that a function might get closer to over time.
Using Infinity
A function at infinity just means that we are evaluating a limit as x gets closer to infinity or negative infinity: lim┬(x→∞)〖f(x)〗 or lim┬(x→-∞)〖f(x)〗. In general, this can be done by simply plugging in infinity as if it were a number and seeing what result follows.
Example 1: Evaluate the infinite limits of x
Solution:
Left limit: lim┬(x→-∞) x= -∞
Right limit:lim┬(x→∞) x=∞
Example 2: Identify the end behavior of -x^2+3
Solution:
Left limit: lim┬(x→-∞)-x^2+3=-(-∞)^2+3=-∞+3=-∞
Right limit:lim┬(x→∞)-x^2+3=-(∞)^2+3=-∞+3=-∞
Example 3: What happens to √(x+4) as x gets larger?
Solution:
lim┬(x→∞) √(x+4)=√(∞+4)=√∞=∞
Comparing Infinities
There are few cases where plugging in is simply not good enough. Any situation in which there is a comparison between infinites will result in a problem we cannot tackle given previous methods.
Example 4: Evaluate the infinite limits of (x+1)/(2x-3)
Solution:
Left limit: lim┬(x→-∞) (x+1)/(2x-3)= (-∞+1)/(2(-∞)-3)=(-∞)/(-∞). This limit is indeterminate.
Right limit:lim┬(x→∞) (x+1)/(2x-3)=(∞+1)/(2(∞)-3)=∞/∞. This limit is indeterminate.
In this case, the method to evaluate this limit involves simplifying the equation in question by dividing the x’s out. This method works by identifying the largest exponent present and multiplying by 1/xexponent on top and bottom of the fraction. This has the effect of leaving only the largest x-value and therefore only the largest infinity. It is helpful to recall that 1/∞=0
Let’s finish Example 4:
Left limit: lim┬(x→-∞) (x+1)/(2x-3)= lim┬(x→-∞) (x+1)/(2x-3)∙(1⁄x)/(1⁄x)=lim┬(x→-∞) (1+1/x)/(2-3/x)=(1+1/(-∞))/(2-3/(-∞))=1/2
Right limit:lim┬(x→∞) (x+1)/(2x-3)=lim┬(x→∞) (x+1)/(2x-3)∙(1⁄x)/(1⁄x)=lim┬(x→∞) (1+1/x)/(2-3/x)=(1+1/∞)/(2-3/∞)=1/2
Here is a second example:
Example 5:
Solution:
This final situation will be expanded on later, once we have talked more about rates of changes.