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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 1

Lesson 19: Unbounded limits (vertical asymptotes)- Infinite limits intro
- Connecting limits at infinity notation and graph
- Infinite limits: graphical
- Analyzing unbounded limits: rational function
- Analyzing unbounded limits: mixed function
- Infinite limits: algebraic
- Visually determining vertical asymptotes (old)

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# Visually determining vertical asymptotes (old)

An old video where Sal identifies all the vertical asymptotes of a function given graphically. Created by Sal Khan.

## Want to join the conversation?

- for limits at infinity where f(x) is unbounded- how do we determine (not visually) if the limit is infinity or negative infinity?(7 votes)
- You can get a vertical asymptote when you have a rational function where your variable is in the denominator, and it is possible to set that variable such that the denominator evaluates to zero. Think about it, as you pick values for the variable that get the denominator closer and closer to zero, the bigger the magnitude of the function -- the value of the function is going to go to positive or negative infinity as the denominator goes to zero. This is a vertical asymptote.

Thus the function f(x) = 1/(x^2 + 4x + 3) has two vertical asymptotes, one at x = -1 and one at x = -3. As these are the two values of x that cause the denominator to get to zero. How do I know that? The denominator can be factored: f(x) = 1/((x + 1)(x + 3)

Now it is easy to see that setting x to -1 or -3 causes the denominator to go to zero.

There is one caveat. If the denominator contains a factor that is also in the numerator, the x value that would cause that factor to be zero, and thus make the whole denominator be zero will NOT cause a vertical asymptote, it will cause a hole in the function. Example:

Let f(x) = (x^2 + 4x + 3)/(x + 1). I can factor this to:

f(x) = (x + 1)(x + 3)/((x + 1). Algebraically I can cancel the (x + 1) factors from the numerator and denominator and end up with a linear function: g(x) = (x + 3). Because I can algebraically cancel the (x + 1) factor from the numerator and denominator, setting x to -1 to make that factor zero will NOT cause a vertical asymptote in the original function, it will cause a hole in the function.(17 votes)

- What does f(x) is unbounded mean?

Thanks(6 votes)- In a domain
**D**, a function f is bounded when a real number M exists to make`|f(x)|≤M`

If M is not defined (i.e. infinite), then f(x) is unbounded.

For example, sin x is bounded because |sin x |≤1, while 1 is its upper bound and -1 is its lower bound.

In this video, the function can be made bounded if the domain is, for example, (2, 8) or (−3, 1).(7 votes)

- let's say we defined x=2 to have f(x)= -4. would x=2 still be an asymptote? (considering that approaching x=2 from the left side will bring us to infinity)(5 votes)
- Yes. We can define a vertical asymptote of a function f(x) to occur at x = a if a one-sided limit of f(x) as x-->a is positive or negative infinity (if it behaves that way from both sides of a, that's okay too). But as always, the limit doesn't care at all about what happens
**at**x = a. So if the function happens to be defined as`f(2) = -4`

there but blows up infinitely on one side, it would still be considered a vertical asymptote at x = 2. Hope that helps!(5 votes)

- Does a LINEAR function have an assymptote? It self?(4 votes)
- Yes; a function
`ƒ`

of the form`ƒ(x) = ax + b`

is its own asymptote.(1 vote)

- What is an unbounded limit?(3 votes)
- Unbounded limit is infinite limit. This is when x->a, f(x)-> ±∞. Such is an example of a vertical asymptote where f(x) -> ±∞. This is called unbounded limit or infinite limit. Though we call it infinite limit. This is a non-existing limit or the limit does not exist (dne). For the limit to exist, x->a, f(x)>#. Since ±∞ is not a #, so its limit dne. But if it dne, why do we say it is infinite limit? Though it dne, we know its end-behavior is approaching ±∞. So instead of saying it dne, it's more accurate to say its end-behavior.(3 votes)

- Hi! I liked the video; I found it mostly helpful for understanding the concepts. However, I do have one question: the video is saying that the limit of a graph as it approaches an asymptote is positive or negative infinity, but why isn't the asymptote the limit? Isn't a limit the phenomenon of which a graph is approaching a certain point?(2 votes)
- The limit is the value that the function approaches, which in this case if f(x) or y. The asymptote is the line representing the value that the input of the function, or x, is approaching. As x approaches negative four, y approaches infinity. Since y is the function, infinity is the limit at that input.(4 votes)

- In the video question is the limit as x -> -4 defined to be infinity but not at x=2?(2 votes)
- Both the left hand and right hand limits as x -> -4 go to infinity, whereas for x -> 2, the left hand limit goes to infinity, but the right hand limit goes to -4, so there is no limit as x->2.

You might want to review this section:

https://www.khanacademy.org/math/differential-calculus/limits-topic/calculus-estimating-limits-graph/v/one-sided-limits-from-graphs(1 vote)

- Can we find a function's asymptote algebraically?(1 vote)
- A vertical asymptote occurs where the function is undefined (e.g., the function is y=A/B, set B=0). A horizontal asymptote (or oblique) is determined by the limit of the function as the independent variable approaches infinity and negative infinity.

Algebraically, there are also a couple rules for determining the horizontal (or oblique asymptote).

If the degree of the numerator is higher than the denominator, an oblique asymptote occurs. When the numerator and denominator have the same degree, the horizontal asymptote is the fraction of the leading coefficients of highest degree term. If the degree of the denominator is less than the degree of the numerator, the function will have a horizontal asymptote as y=0 (because the limit of the functions approaches 0).(2 votes)

- Anyone else hear some kind of alarm going off in the background??(1 vote)
- Why/how is it enough to conclude if something has a vertical asymptote just by looking at the unbounded one sided limit? And is a graph unbounded if it has a finite value somewhere on the asymptote?(1 vote)

## Video transcript

- [Instructor] Given the graph of y equals f of x pictured below, determine the equations of
all vertical asymptotes. Let's see what's going on here. So it looks like interesting
things are happening at x equals negative
four and x equals two. At x equals negative
four, as we approach it from the left the value of the function just becomes unbounded right over here. Looks like as we approach
x equals negative four from the left, the value of
our function goes to infinity. Likewise, as we approach
x equals negative four from the right, it looks like our, the value of our function
goes to infinity. So, I'd say that we definitely
have a vertical asymptote at x equals negative four. Now, let's look at x equals two. As we approach x equals two from the left, the value of our function once
again approaches infinity, or it becomes unbounded. Now, from the right, we
have an interesting thing. If we look at the limit from
the right right over here, it looks like we're
approaching a finite value. As we approach x equals
two from the right, it looks like we're approaching f of x is equal to negative four. But, just having a one-sided
limit that is unbounded, is enough to think about
this as a vertical asymptote. The function is not
defined right over here, and as we approach it from just one side, we are becoming unbounded. It looks like we're approaching infinity or negative infinity, so that by itself, this unbounded left hand limit,
or left-side limit by itself is enough to consider x equals
two a vertical asymptote. So, we can say that there's
a vertical asymptote at x equals negative
four and x equals two.