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In the previous section we saw limits that were infinity and
it’s now time to take a look at limits at infinity. By limits at infinity we mean one of the
following two limits.
In other words, we are going to be looking at what happens
to a function if we let x get very
large in either the positive or negative sense.
Also, as well soon see, these limits may also have infinity as a value.
For many of the limits that we’re going to be looking at we
will need the following facts.
Fact 1
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1. If r is a positive rational number and c is any real number then,
2. If r is a positive rational number, c is any real number and xr is defined for then,
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The first part of this fact should make sense if you think
about it. Because we are requiring we know that xr will stay in the denominator. Next as we increase x then xr will
also increase. So, we have a constant
divided by an increasingly large number and so the result will be increasingly
small. Or, in the limit we will get
zero.
The second part is nearly identical except we need to worry
about xr being defined for
negative x. This condition is here to avoid cases such as
. If this r
were allowed then we’d be taking the square root of negative numbers which
would be complex and we want to avoid that at this level.
Note as well that the sign of c will not affect the answer.
Regardless of the sign of c
we’ll still have a constant divided by a very large number which will result in
a very small number and the larger x
get the smaller the fraction gets. The
sign of c will affect which direction
the fraction approaches zero (i.e.
from the positive or negative side) but it still approaches zero.
To see the proof of this fact see the Proof of Various Limit Properties
section in the Extras chapter.
Let’s start the off the examples with one that will lead us
to a nice idea that we’ll use on a regular basis about limits at infinity for
polynomials.
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Example 1 Differentiate
each of the following functions.
(a) [Solution]
(b) [Solution]
Solution
(a)
Our first thought here is probably to just “plug” infinity
into the polynomial and “evaluate” each term to determine the value of the
limit. It is pretty simple to see what
each term will do in the limit and so this seems like an obvious step,
especially since we’ve been doing that for other limits in previous
sections.
So, let’s see what we get if we do that. As x
approaches infinity, then x to a
power can only get larger and the coefficient on each term (the first and
third) will only make the term even larger.
So, if we look at what each term is doing in the limit we get the
following,
Now, we’ve got
a small, but easily fixed, problem to deal with. We are probably tempted to say that the
answer is zero (because we have an infinity minus an infinity) or maybe (because we’re subtracting two
infinities off of one infinity).
However, in both cases we’d be wrong.
This is one of those indeterminate
forms that we first started seeing in a previous section.
Infinities just
don’t always behave as real numbers do when it comes to arithmetic. Without more work there is simply no way to
know what will be and so we really need to be careful
with this kind of problem. To read a
little more about this see the Types of
Infinity section in the Extras chapter.
So, we need a
way to get around this problem. What
we’ll do here is factor the largest power of x out of the whole polynomial as follows,
If you’re not
sure you agree with the factoring above (there’s a chance you haven’t really
been asked to do this kind of factoring prior to this) then recall that to
check all you need to do is multiply the back through the parenthesis to verify it
was done correctly. Also, an easy way
to remember how to do this kind of factoring is to note that the second term
is just the original polynomial divided by . This will always work when factoring a
power of x out of a polynomial.
Next, from our
properties of limits we know that the limit of a product is the product of
the limits so we can further write this as,
The first limit
is clearly infinity and for the second limit we’ll use the fact above on the
last two terms and so we’ll arrive at the following value of the limit,
Note that while
we can’t give a value for ,
if we multiply an infinity by a constant we will still have an infinity no
matter how large or small the constant is.
The only thing that we need to be careful of is signs. If the constant had been negative then we’d
have gotten negative infinity for a value.
[Return to Problems]
(b)
We’ll work this part much quicker than the previous
part. All we need to do is factor out
the largest power of t to get the
following,
Remember that
all you need to do to get the factoring correct is divide the original
polynomial by the power of t we’re
factoring out, in this case.
Now all we need
to do is take the limit of the two terms.
In the first don’t forget that since we’re going out towards and we’re raising t to the 5th power that the limit will be negative
(negative number raised to an odd power is still negative). In the second term well again make heavy
use of the fact above.
So, taking the
limits of the two terms gives,
Note that
dividing an infinity (positive or negative) by a constant will still give an
infinity.
[Return to Problems]
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Okay, now that we’ve seen how a couple of polynomials work
we can give a simple fact about polynomials in general.
Fact 2
What this fact is really saying is that when we go to take a
limit at infinity for a polynomial then all we need to really do is look at the
term with the largest power and ask what that term is doing in the limit since
the polynomial will have the same behavior.
You can see the proof in the Proof of Various Limit
Properties section in the Extras chapter.
Let’s now move into some more complicated limits.
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Example 2 Evaluate
both of the following limits.
Solution
First, the only difference between these two is that one
is going to positive infinity and the other is going to negative
infinity. Sometimes this small
difference will affect then value of the limit and at other times it won’t.
Let’s start with the first limit and as with our first set
of examples it might be tempting to just “plug” in the infinity. Since both the numerator and denominator
are polynomials we can use the above fact to determine the behavior of
each. Doing this gives,
This is yet another indeterminate form. In this case we
might be tempted to say that the limit is infinity (because of the infinity
in the numerator), zero (because of the infinity in the denominator) or -1
(because something divided by itself is one).
There are three separate arithmetic “rules” at work here and without
work there is no way to know which “rule” will be correct and to make matters
worse it’s possible that none of them may work and we might get a completely
different answer, say to pick a number completely at random.
So, when we have a polynomial divided by a polynomial
we’re going to proceed much as we did with only polynomials. We first identify the largest power of x in the denominator (and yes, we only
look at the denominator for this) and we then factor this out of both the
numerator and denominator. Doing this
for the first limit gives,
Once we’ve done this we can cancel the from both the numerator and the denominator
and then use the Fact 1 above to take the limit of all the remaining
terms. This gives,
In this case the indeterminate form was neither of the
“obvious” choices of infinity, zero, or -1 so be careful with make these
kinds of assumptions with this kind of indeterminate forms.
The second limit is done in a similar fashion. Notice however, that nowhere in the work
for the first limit did we actually use the fact that the limit was going to
plus infinity. In this case it doesn’t
matter which infinity we are going towards we will get the same value for the
limit.
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In the previous example the infinity that we were using in
the limit didn’t change the answer. This
will not always be the case so don’t make the assumption that this will always
be the case.
Let’s take a look at an example where we get different
answers for each limit.
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Example 3 Evaluate
each of the following limits.
Solution
The square root in this problem won’t change our work, but
it will make the work a little messier.
Let’s start with the first limit. In this case the largest power of x in the denominator is just an x.
So we need to factor an x
out of the numerator and the denominator.
When we are done factoring the x
out we will need an x in both of
the numerator and the denominator. To
get this in the numerator we will have to factor an x2 out of the square root so that after we take the
square root we will get an x.
This is probably not something you’re used to doing, but
just remember that when it comes out of the square root it needs to be an x and the only way have an x come out of a square is to take the
square root of x2 and so
that is what we’ll need to factor out of the term under the radical. Here’s the factoring work for this part,
This is where we need to be really careful with the square
root in the problem. Don’t forget that
Square roots are ALWAYS positive and so we need the
absolute value bars on the x to
make sure that it will give a positive answer. This is not something that most people
every remember seeing in an Algebra class and in fact it’s not always given
in an Algebra class. However, at this
point it becomes absolutely vital that we know and use this fact. Using this fact the limit becomes,
Now, we can’t just cancel the x’s. We first will need to
get rid of the absolute value bars. To
do this let’s recall the definition of absolute value.
In this case we are going out to plus infinity so we can
safely assume that the x will be
positive and so we can just drop the absolute value bars. The limit is then,
Let’s now take a look at the second limit (the one with
negative infinity). In this case we
will need to pay attention to the limit that we are using. The initial work will be the same up until
we reach the following step.
In this limit we are going to minus infinity so in this
case we can assume that x is
negative. So, in order to drop the
absolute value bars in this case we will need to tack on a minus sign as
well. The limit is then,
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So, as we saw in the last two examples sometimes the
infinity in the limit will affect the answer and other times it won’t. Note as well that it doesn’t always just
change the sign of the number. It can on
occasion completely change the value.
We’ll see an example of this later in this section.
Before moving on to a couple of more examples let’s revisit
the idea of asymptotes that we first saw in the previous section. Just as we can have vertical asymptotes
defined in terms of limits we can also have horizontal asymptotes defined in
terms of limits.
Definition
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The function f(x)
will have a horizontal asymptote at y=L
if either of the following are true.
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We’re not going to be doing much with asymptotes here, but
it’s an easy fact to give and we can use the previous example to illustrate all
the asymptote ideas we’ve seen in the both this section and the previous
section. The function in the last
example will have two horizontal asymptotes.
It will also have a vertical asymptote. Here is a graph of the function showing these.
Let’s work another couple of examples involving of rational
expressions.
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Example 4 Evaluate
each of the following limits.
Solution
Let’s do the first limit and in this case it looks like we
will factor a z3 out of
both the numerator and denominator.
Remember that we only look at the denominator when determining the
largest power of z here. There is a larger power of z in the numerator but we ignore
it. We ONLY look at the denominator
when doing this! So doing the
factoring gives,
When we take the limit we’ll need to be a little
careful. The first term in the
numerator and denominator will both be zero.
However, the z3
in the numerator will be going to plus infinity in the limit and so the limit
is,
The final limit is negative because we have a quotient of
positive quantity and a negative quantity.
Now, let’s take a look at the second limit. Note that the only different in the work is
at the final “evaluation” step and so we’ll pick up the work there.
In this case
the z3 in the numerator
gives negative infinity in the limit since we are going out to minus infinity
and the power is odd. The answer is
positive since we have a quotient of two negative numbers.
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