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### Course: Differential Calculus > Unit 1

Lesson 14: Infinite limits- Introduction to infinite limits
- Infinite limits and asymptotes
- Connecting limits at infinity notation and graph
- Infinite limits: graphical
- Analyzing unbounded limits: rational function
- Analyzing unbounded limits: mixed function
- Infinite limits: algebraic

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# Analyzing unbounded limits: rational function

Sal analyzes the behavior of f(x)=-1/(x-1)² around its asymptote at x=1.

## Want to join the conversation?

- can we say in that case the limit as x approach 1 will be negative infinity.. because the left and the right limit are the same ??(7 votes)
- We can (and often do) say that, but to be completely correct we should say something like "the limit as x approaches 1 diverges towards negative infinity".

The reason we aren't supposed to say a limit will be (negative) infinity, is that infinity is**not**a*real number*– it might be better thought of as a direction ...

If you are interested, there is a whole area of mathematics devoted to infinities!(25 votes)

- i cant believe the last comment on this video was 3 years ago.... well i better update that ig(6 votes)
- let me see...i'm going through and answering questions..nut sure this is a question...(8 votes)

- LETS SAY... x WAS! EQUAL TO 1 dun dun duoooooon.

Wa will happon(1 vote)- If x was equal to one the function will be undefined. The whole reason while we use limits is to approximate the values of functions at undefined points so making it equal to one negates that concept. This graph shows the graph of the function given in this graph:https://www.desmos.com/calculator/ou6ywa7vsi(2 votes)

- Throughout the video, is it incorrect for it to be saying that the limit of the function = positive or negative infinity? I ask because thinking the way it talks in this video made me think that both limits talked about in this video exist, which Sal says is wrong:

https://www.khanacademy.org/math/differential-calculus/dc-limits/dc-point-continuity/v/continuity-at-a-point-graphically(1 vote)- Unbounded limits don't exist; however, they are different from limits such as a_n = (-1)^n ; this sequence doesn't have a limit merely because it is alternating between 1 & -1, though its absolute value stays at 1. Unbounded limits aren't oscillating - they keep getting bigger or smaller. So we define infinity & - infinity to represent that. Technically they aren't "real numbers" but they are apart of the extended real number system. A reason as to why the limits can't exist is because consider 1 = x*1/x (x > 0) as x approaches 0 from the right. If the limit existed we could write lim x * 1/x = lim x * lim 1/x = 0 * (infinity) = 0. But the limit is clearly 1. So saying the limit doesn't exist is just a reminder we can't use limit properties to pull apart operations.(0 votes)

- how can i know the limit as x approaches 1 in the above function ?(0 votes)
- - infinity.

That's exactly what was found out in the video, since both the one-sided limits = - infinity, the two-sided limit also is - infinity(0 votes)

- is there any other way to do it except the estimating tabls way(0 votes)
- Yes, these techniques are covered in the rest of the limits playlist.(2 votes)

- Can I put a function and then you give me the answers of its limits step by step?(0 votes)
- Would you want me to find a limit here in the answer section? I would be glad to help.(4 votes)

- Hi do you like to do math(0 votes)
- How can you put a sign on infinity? I can prove you can't given 1/∞=0.

x=x

x=x*x/x

x=(x^2)(1/x)

|x|=|x^2||1/x|

|x|=(x^2)|1/x|

|x|/x=|1/x|(x^2)/x

|x|/x=|1/x|x

|x|/x=|1/x|/(1/x)

Now substitute +-∞. You get:

∞/(+-∞)=|1/∞|/(1/+-∞)

We know that 1/∞=0, so 1/0=∞, so:

∞/(+-∞)=0/+-0

∞/(+-∞)=0/0

∞/(+-∞)=(1/∞)/(1/∞)

∞/(+-∞)=∞/∞

(+-∞)/∞=∞/∞

+-∞=∞

Positive and negative infinity are both actually just plain old infinity. Right?(0 votes)- Limits and arithmetic are different.

You need to take into account if limit is indeterminable as well.(2 votes)

## Video transcript

- [Voiceover] Let f of x
be equal to negative one over x minus one squared. Select the correct description
of the one-sided limits of f at x equals one. And so we can see, we
have a bunch of choices where we're approaching x
from the right-hand side and we're approaching x
from the left-hand side. And we're trying to figure
out do we get unbounded on either of those, in the positive, towards positive infinity
or negative infinity. And there's a couple of ways to tackle it. The most straightforward, well, let's just consider each
of these separately. So we can think about the limit of f of x as x approaches one from
the positive direction and limit of f of x as x approaches one, as x approaches one
from the left-hand side. This is from the right-hand side. This is from the left-hand side. So I'm just gonna make a
table and try out some values as we approach, as we approach
one from the different sides, x, f of x, and I'll do
the same thing over here. So, we are going to have
our x and have our f of x and if we approach one from
the right-hand side here, that would be approaching one from above, so we could try 1.1, we could try 1.01. Now f of 1.1 is negative one
over 1.1 minus one squared. So see this denominator here
is going to be .1 squared. So this is going to be,
this is going to be 0.01, and so this is going to be negative 100. So let me just write that down. That's going to be negative 100. So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared. Well, then this denominator
this is going to be, this is the same thing as 0.01 squared, which is the same thing
as 0.0001, 1/10000. And so the negative one 1/10000 is going to be negative 10,000. So, let's just write that
down, negative 10,000. And so this looks like, as we get closer, 'cause notice, as I'm going
here I am approaching one from the positive direction, I'm getting closer and
closer to one from above and I'm going unbounded
towards negative infinity. So this looks like it
is negative infinity. Now we can do the same thing
from the left-hand side. I could do 0.9, I could do 0.99. Now 0.9 is actually also
going to get me negative 100 'cause 0.9 minus one is
going to be negative .1 but then when you square
it the negative goes away so you get a .01 and then
one divided by that is 100 but you have the negative,
so this is also negative 100. And if you don't follow those
calculations, I'll do it, let me do it one more time
just so you see it clearly. This is going to be negative one over, so now I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from
below from the left-hand side. So this is going to be
0.99 minus one squared. Well, 0.99 minus one is, is
going to be negative 1/100, so this is going to be
negative 0.01 squared. When you square it the negative goes away and you're left with 1/10000. So this is going to be 0.0001 and so when you evaluate
this you get 10,000. So that, or sorry, you
get negative 10,000. So in either case,
regardless of which direction we approach from, we are
approaching negative infinity. So that is this choice right over here. Now there's other ways you
could have tackled this if you just look at, kind of, the structure of this expression here, the numerator is a constant, so that's clearly always
going to be positive. Let's ignore this negative
for the time being. That negative's out front. This numerator, this one is
always going to be positive. Down here, we're taking at x equals one, while this becomes zero
and the whole expression becomes undefined, but as we approach one, x minus one could be positive or negative as we see over here, but
then when we square it, this is going to become positive as well. So the denominator is going to be positive for any x other than one. So positive divided by
positive is gonna be positive but then we have a negative out front. So this thing is going to be negative for any x other than one, and it's actually not
defined at x equals one. And so you could, from
that, you could deduce, well, okay then, we can
only go to negative infinity there's actually no way
to get positive values for this function.