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# Limits and End Behavior - Problem 2

FREE###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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## tulisamarth@hotmail.com · 2 weeks ago

Mr. Prokup has significantly simplified the essential understanding of continuity, limits, and end behavior and providently is innovative with his example and characterizing the evidence and proof behind the limits of the functions in a very simplisitic and non-exorbitant way. I really want to thank him because I am in 7th grade and I'm doing an introduction to Calculus AB right now and this is really helping to expand my interest for calculus now.