# Limits and End Behavior - Problem 1

FREEEnd behavior refers to finding out what happens to a function as x goes to positive or negative infinity. The trick to calculating limits going toward any infinity is to find what the largest power of x in the denominator is, then to multiply both the numerator and denominator by 1 over that power. For example, for a function f(x)=(x^3+7)/(x^2-x+1), multiply both the numerator and denominator by 1/x^2. As x goes to positive or negative infinity, 1/x, 1/x^2, 1/x^3, and so on, become 0. This is because when a fraction's denominator is increasing, the fraction gets smaller (this makes sense with numbers: 1/4 is less than 1/2 because 4 is greater than 2). So, for any terms in your function that are a number over a power of x, these will become 0. Going back to our example, when both the numerator and denominator are multiplied by 1/x^2, f(x)=(x+7/x^2)/(1-1/x+1/x^2). All of the terms dividing a power of x will become 0, so f(x) as x approaches infinity will become x/1. Then, you can evaluate the limit as usual. Recall that the degree of a polynomial is the largest power of x (so a polynomial x^7+9x^4+2 is a polynomial of degree 7). If the degree of the numerator is the same as the degree of the denominator, repeat the same procedure as described. In this case, the horizontal asymptote a line y=b, where b is the coefficient of the numerator divided by the coefficient of the denominator. When x approaches negative infinity, repeat the same procedure, but when examining the polynomial after multiplying the numerator and denominator by 1 over the largest power of x, remember that odd powers of x will make the polynomial negative, and even powers will make the polynomial positive. So f(x)=4x^3 as x approaches negative infinity will become negative infinity, but g(x)=4x^4 as x approaches negative infinity will become positive infinity.

Let’s take a look at an example. If g(x) is 25x² over x² plus 9, let’s compute a, the limit as x approaches the infinity of g(x), and then b for limit as x approaches –infinity of g(x). This is going to require a trick. So first of all I’m going to look at a function g(x), 25x² over x² plus 9.

As x goes to infinity, you can see that the both numerator and denominator are getting very large. So how do we determine what happens to the quotient? You need to use a trick. The trick is to multiply by 1 over x² on top and bottom.

Now how do I pick that 1 over x²? You look at the denominator and the highest power of x in the denominator determines what you have to multiply top and bottom by. It's 1 over that power of x. And when you do that, take a look of what we get. 25x² times 1 over x² is just 25.

X² times 1 over x² is 1, 9 times 1 over x² is 9 over x². This function is exactly the same as this one. We multiply by 1 but it’s a lot easier to see in this case what happens as x goes to infinity.

As x goes to infinity, the denominator gets huge and this is going to get very small. This is going to go to 0. So limit as x approaches infinity g(x), is the limit as x approaches infinity of this thing; 25 over 1 plus 9 over x². And that’s 25 over 1 plus 0 which is just 25.

Now I can use the exact same trick on this limit. This is going to be equal to the limit as x approaches negative infinity and g(x) is still 25 over 1 plus 9 over x². And it works exactly the same as x approaches negative infinity the denominator is getting huge.

So this thing is going 0. The limit is 25 over 1 plus 0. 25. What this indicates is that y equals 25 is a horizontal asymptote. Let’s take a look at the graph of this function.

So here’s the horizontal asymptote; y equals 25. And here’s the graph of the function ;f(x) this is g(x) equals 25x² over x² plus 25 or plus 9 rather. So what happens is x gets huge, the graph goes closer and closer to the value 25, and it does that as x approaches negative infinity as well. So y equals 25 is a horizontal asymptote.

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