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Current time:0:00Total duration:4:07

AP.CALC:

LIM‑2 (EU)

, LIM‑2.D (LO)

, LIM‑2.D.3 (EK)

, LIM‑2.D.4 (EK)

, LIM‑2.D.5 (EK)

So we have f of x equaling 4x
to the fifth minus 3x squared plus 3, all of that over
6x to the fifth minus 100x squared minus 10. Now, what I want
to think about is-- what is the limit of f of
x, as x approaches infinity? And there are several ways
that you could do this. You could actually try to plug
in larger and larger numbers for x and see if it seems to
be approaching some value. Or you could reason
through this. And when I talk about
reasoning through this, it's to think about the
behavior of this numerator and denominator as x gets
very, very, very large. And when I'm talking about
that, what I'm saying is, as x gets very, very
large-- let's just focus on the numerator. As x gets very, very
large, this term right over here in the numerator--
4x to the fifth-- is going to become a much,
much more significant than any of these other things. Something squaring gets large. But something being
raised to the fifth power gets raised that
much, much faster. Similarly, in the denominator,
this term right over here-,, the highest degree
term-- 6x to the fifth-- is going to grow much, much,
much faster than any of these other terms. Even though this has 100 as a
coefficient or a negative 100 as a coefficient, when you take
something to the fifth power, it's going to grow so much
faster than x squared. So as x gets very,
very, very large, this thing is going
to approximate 4x to the fifth over 6x to the
fifth for a very large, large x Or we could say as x
approaches infinity. Now, what could this
be simplified to? Well, you have x to the fifth
divided by x to the fifth. These are going
to grow together. So these you can think
of them as canceling out. And so you are left with 2/3. So what you could say
is-- the limit of f of x, as x approaches infinity,
as x gets larger and larger and larger, all of
these other terms aren't going to
matter that much. And so it's going
to approach 2/3. Now, let's look at
the graph and see if that actually makes sense. What we're actually
saying is that we have a horizontal asymptote
at y is equal to 2/3. So lets look at the graph. So right here is the graph. Got it from Wolfram Alpha. And we see, indeed, as x gets
larger and larger and larger, f of x seems to be approaching
this value that looks right at around 2/3. So it looks like we have
a horizontal asymptote right over here. Let me draw that a
little bit neater. We have a horizontal
asymptote right at 2/3. So let me draw it
as neatly as I can. So this right over here
is y is equal to 2/3. The limit as x gets
really, really large, as it approaches infinity, y
is getting closer and closer and closer to 2/3. And when we just look
at the graph here, it seems like the same
thing is happening from the bottom direction, when
x approaches negative infinity. So we could say the
limit of f of x, as x approaches
negative infinity, that also looks like it's 2/3. And we can use the
exact same logic. When x becomes a very,
very, very negative number, as it becomes
further and further to the left on the number
line, the only terms that are going to matter are
going to be the 4x to the fifth and the 6x to the fifth. So this is true
for very large x's. It's also true for
very negative x's. So we could also say, as x
approaches negative infinity, this is also true. And then, the x to the fifth
over the x to the fifth is going to cancel out. These are the dominant terms. And we're going to
get it equaling 2/3. And once again, you see
that in the graph here. We have a horizontal asymptote
at y is equal to 2/3. We take the limit of f of
x as x approaches infinity, we get 2/3. And the limit of f of x as x
approaches negative infinity is 2/3. So in general,
whenever you do this, you just have to
think about what terms are going to
dominate the rest? And focus on those.