##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Infinite Limits; Vertical Asymptotes - 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.

In order to see if there is a vertical asymptote in the graph of a rational function, find where the denominator is 0. Remember, when the denominator of a function is 0, it is undefined because you cannot divide by 0. When the function is undefined, there is a vertical asymptote. In order to see if f(x) goes toward positive or negative infinity as x approaches this point, look at the values of f(x) as x approaches the point from the left and the right. When f(x) gets larger or more and more positive, f(x) approaches positive infinity. When f(x) gets smaller or more and more negative, f(x) approaches negative infinity. It is possible for f(x) to approach positive infinity from one side and negative infinity from the other, or for f(x) to approach positive or negative infinity from both sides.

Let's take a look at another one-sided limit problem. We have g of x equals 24x over x² minus 9. I want to find the limit as approaches 3 from the right of g of x. You can see that 3 is going to be an interesting number, because the denominator is 0 when x equals 3.

So let me take a closer look at the function g of x equals 24x over x² minus 9. What I like to do is I like to algebraically separate the part that has a problem with x equals 3. That's one of the factors of the denominator. This is 24x over x plus 3, x minus 3. That's the problem right there. If you plug in 3, you get a 0 right here. So let's kind of get that part out.

We have 24x over x plus 3, times 1 over x minus 3. So we've isolated the problem part. Let's see what happens to each of these pieces as x approaches 3. As x approaches 3 from the right, the 24x over x plus 3 part, goes to, well 24 times 3 is 72. 3 plus 3 is 6. So this part is going to 12.

What about the other part? 1 over x minus 3. X is approaching 3 from the right. So think about numbers that are a little bit bigger than 3 like 3.1 or 3.01. If I plug in 3.01, I get 3.01 minus 3.01, 1 over 0.01 is a 100. So this is going up to positive infinity. So this thing is headed towards 12 times infinity. The whole thing is going to infinity.

Let's look at the other one-sided limit. Limit as x approaches 3 from the left. Same analysis, I have that g of x is still 24x over x plus 3 times 1 over x minus 3. I want to see what happens this time as x approaches 3 from the left. So again the 24x over x plus 3. The top part is still headed towards 72. As x approaches 3, 72. The bottom part is headed towards 6. So this is still going to 12. What about the 1 over x minus 3?

As x approaches 3 from the left, think about numbers that are a little to the left of 3, like 2.99. 2.99 minus 3 is like -0.01. 1 over -0.01 is -100. This is going down to negative infinity. So the product is headed towards 12 times negative infinity. It's headed towards negative infinity. So this limit is negative infinity.

Now what does that mean about vertical asymptotes? Well, the fact is we have more than we need to prove that there's a vertical asymptote at x equals 3. As x approaches 3 from the left, the function is going to negative infinity. From the right it's going to positive infinity. So x equals 3 is definitely a vertical asymptote.

Please enter your name.

Are you sure you want to delete this comment?

###### 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!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (2)

Please Sign in or Sign up to add your comment.

## ·

Delete

## Rumbie · 1 month, 4 weeks ago

Very helpful thank you

## Rumbie · 1 month, 4 weeks ago

Very helpful thank you