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## Connecting infinite limits and vertical asymptotes

# Analyzing unbounded limits: rational function

AP.CALC:

LIM‑2 (EU)

, LIM‑2.D (LO)

, LIM‑2.D.1 (EK)

, LIM‑2.D.2 (EK)

## 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.

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