# Infinite Limits; Vertical Asymptotes - Problem 1

By looking at the limit as x approaches a certain point, you can find where there are vertical asymptotes. When the limit is undefined, look at the limit as x approaches the point from the left and from the right, and what happens as x gets closer and closer to that point. By calculating values of f(x) as x gets closer to the undefined point, you can see if it goes toward positive infinity (if f(x) gets more and more positive) or negative infinity (if f(x) gets more and more negative).

Let's take a look at another example. Consider the function h(f) equals 2x² minus x minus 9 over x plus 2. I want to find the limit as x approaches -2 from the right of h(x). Now h(x) which equals 2x² minus x minus 9 over x plus 2. I can separate the top, and bottom halves of this function. I can write it as 2x² minus x minus 9, times 1 over x plus 2.

The reason I'm doing that is that, this is the problematic part. As x approaches -2, this is the part that's undefined. So I'm separating that out. Now let's look at what happens as x approaches -2 to these two pieces.

As x approaches -2, what happens to the quadratic part? 2x² minus x minus 9. That's going to approach 2 times 4, -2² is 4. That's 8, minus -2, so +2, minus 9. So that part approaches 1. This thing is going towards 1.

What about this guy? 1 over x plus 2. As x approaches -2 from the positive side, this thing is going to 0. It's important to know whether it's going to go 0 from the positive, or from the negative. Let's take a number that's close to -2, but on the positive side like -1.9. -1.9 plus 2 is going to give me 0.1. 1 over 0.1 is 10. So it's going from the positive side to 0. That means this quantity is going to positive infinity.

So this quantity is going towards 1 times infinity. We conclude that this limit is infinity.

Now let's take a look at the other limit. The limit as x approaches -2 from the left. Same function. I can use the same kind of analysis I did here. I can use the fact that h(x) is this quadratic times 1 over x plus 2. So 2x² minus x minus 9 times 1 over x plus 2. This time, x is going towards -2 from the left. So what's happening to the quadratic? It's really the same thing. It's not sensitive to whether x is going to -2 from the right or left. 2x² minus 2 minus 9 is going towards, again, 2 times 4, 8, minus -2, so plus 2, minus 9 which is 1.

What about this piece? So x is approaching -2 from the left. So take a number that's a little bit to the left of -2 like -2.1. -2.1 plus 2 is -0.1, 1 over -0.1 is like -10. This thing is going down to negative infinity. It's important to know whether it's going to positive or negative infinity, because that determines the fate of this function.

This thing is going towards 1 times negative infinity. So it's going towards negative infinity. Now remember, about vertical asymptotes, if you have a limit as x approaches -2, from the right or left, it's going to plus or minus infinity, then that number is going to give you a vertical asymptote. X equals -2 is a vertical asymptote for this function.

## Comments (1)

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## Riley · 2 months, 1 week ago

When you plug in (2.1) for a number left of (2) for the equation (1/x+2), how do you know it's decreasing without bound? I got confused at this part, is it because it yields a negative number?