Limits and End Behavior - Problem 2 2,916 views
With exponential functions, recall that e^x is defined on all x. As x approaches infinity, e^x also approaches infinity. However, as x approaches infinity, e^(-x) (which is the same as 1/e^x) will approach 0. This is because when the denominator of a number gets larger and larger, the number itself gets smaller and smaller. So, using techniques to evaluate limits, you can find the horizontal asymptote of a function involving exponents.
Let’s do a harder example. I have a function h of x equals 5 over 1 plus 4e to the minus x and I want to compute two limits. Limit as x approaches infinity of h of x, the limit as x approaches negative infinity of h of x. And I also want to answer the question whether they are horizontal asymptotes if there are any.
Let’s take a look at this limit first. As x goes to infinity let’s take a look at this piece by piece. Really the only part of this function has a variable is e to the –x. What happens to that as x goes to infinity?. You probably remember the graph of y equals e to the –x looks like this. It's an exponential decay function, and it actually approaches 0 as x approaches infinity. So this thing is going to 0.
What’s 1 plus 4 times that going to? This part is going to 0 so this whole thing is going to 1. And that means that this quotient 5 over 1 plus 4e to the minus x is going towards 5 over 1, 5. And so we have their limit as x approaches infinity of h of x equals 5.
What this means is, that y equals 5 is a horizontal asymptote. Let’s take a look at the limit as x approaches negative infinity. As x approaches negative infinity what happens to that e to the minus x? We are looking back to the graph again.
E to minus x is going up to infinity. So 1 plus 4 times e to the minus x, that’s the denominator of my function, where is that going? Also up to infinity. This is getting big so it’s 4 times that plus 1. But then what happens to 5 over 1 plus 4e to the minus x? The denominator is getting big. So this thing is going to 0.
And I conclude that the limit as x approached negative infinity of h of x is 0 and that means, that this function has y equals 0 as a horizontal asymptote. There not a lot of functions that have two horizontal asymptotes but this one does. Here’s its graph. Y equals 0 and y equals 5 and it kinds of spends its time in between those two horizontal asymptotes. You could see as x approaches negative infinity it approaches the value 0. And as x approaches positive infinity, the graph is going towards 5. So its horizontal asymptotes are y equals 0 and y equals 5.