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In the previous section
we saw that there is a large class of function that allows us to use
to compute limits.
However, there are also many limits for which this won’t work
easily. The purpose of this section is
to develop techniques for dealing with some of these limits that will not allow
us to just use this fact.
Let’s first got back and take a look at one of the first
limits that we looked at and compute its exact value and verify our guess for
the limit.
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Example 1 Evaluate
the following limit.
Solution
First let’s notice that if we try to plug in we get,
So, we can’t just plug in to evaluate the limit. So, we’re going to have to do something
else.
The first thing that we should always do when evaluating
limits is to simplify the function as much as possible. In this case that means factoring both the
numerator and denominator. Doing this
gives,
So, upon factoring we saw that we could cancel an from both the numerator and the
denominator. Upon doing this we now
have a new rational expression that we can plug into because we lost the division by zero
problem. Therefore, the limit is,
Note that this is in fact what we guessed the limit to be.
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On a side note, the 0/0 we initially got in the previous
example is called an indeterminate form. This means that we don’t really know what it
will be until we do some more work.
Typically zero in the denominator means it’s undefined. However that will only be true if the
numerator isn’t also zero. Also, zero in
the numerator usually means that the fraction is zero, unless the denominator
is also zero. Likewise anything divided
by itself is 1, unless we’re talking about zero.
So, there are really three competing “rules” here and it’s
not clear which one will win out. It’s
also possible that none of them will win out and we will get something totally
different from undefined, zero, or one.
We might, for instance, get a value of 4 out of this, to pick a number
completely at random.
There are many more kinds of indeterminate forms and we will
be discussing indeterminate forms at length in the next chapter.
Let’s take a look at a couple of more examples.
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Example 2 Evaluate
the following limit.
Solution
In this case we also get 0/0 and factoring is not really
an option. However, there is still
some simplification that we can do.
So, upon multiplying out the first term we get a little
cancellation and now notice that we can factor an h out of both terms in the numerator which will cancel against
the h in the denominator and the
division by zero problem goes away and we can then evaluate the limit.
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Example 3 Evaluate
the following limit.
Solution
This limit is going to be a little more work than the
previous two. Once again however note
that we get the indeterminate form 0/0 if we try to just evaluate the
limit. Also note that neither of the
two examples will be of any help here, at least initially. We can’t factor and we can’t just multiply
something out to get the function to simplify.
When there is a square root in the numerator or
denominator we can try to rationalize and see if that helps. Recall that rationalizing makes use of the
fact that
So, if either the first and/or the second term have a
square root in them the rationalizing will eliminate the root(s). This might
help in evaluating the limit.
Let’s try rationalizing the numerator in this case.
Remember that to rationalize we just take the numerator
(since that’s what we’re rationalizing), change the sign on the second term
and multiply the numerator and denominator by this new term.
Next, we multiply the numerator out being careful to watch
minus signs.
Notice that we didn’t multiply the denominator out as
well. Most students come out of an
Algebra class having it beaten into their heads to always multiply this stuff
out. However, in this case multiplying
out will make the problem very difficult and in the end you’ll just end up
factoring it back out anyway.
At this stage we are almost done. Notice that we can factor the numerator so
let’s do that.
Now all we need to do is notice that if we factor a
“-1”out of the first term in the denominator we can do some canceling. At that point the division by zero problem
will go away and we can evaluate the limit.
Note that if we had multiplied the denominator out we
would not have been able to do this canceling and in all likelihood would not
have even seen that some canceling could have been done.
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So, we’ve taken a look at a couple of limits in which
evaluation gave the indeterminate form 0/0 and we now have a couple of things
to try in these cases.
Let’s take a look at another kind of problem that can arise
in computing some limits involving piecewise functions.
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Example 4 Given
the function,
Compute the following limits.
(a) [Solution]
(b) [Solution]
Solution
(a)
In this case there really isn’t a whole lot to do. In doing limits recall that we must always
look at what’s happening on both sides of the point in question as we move in
towards it. In this case is completely inside the second interval for
the function and so there are values of y
on both sides of that are also inside this interval. This means that we can just use the fact to
evaluate this limit.
[Return to Problems]
(b)
This part is the real point to this problem. In this case the point that we want to take
the limit for is the cutoff point for the two intervals. In other words we can’t just plug into the second portion because this
interval does not contain values of y
to the left of and we need to know what is happening on
both sides of the point.
To do this part we are going to have to remember the fact
from the section on one-sided limits
that says that if the two one-sided limits exist and are the same then the
normal limit will also exist and have the same value.
Notice that both of the one sided limits can be done here
since we are only going to be looking at one side of the point in question. So let’s do the two one-sided limits and
see what we get.
So, in this case we can see that,
and so since the two one sided limits aren’t the same
doesn’t exist.
[Return to Problems]
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Note that a very simple change to the function will make the
limit at exist so don’t get in into your head that
limits at these cutoff points in piecewise function don’t ever exist.
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Example 5 Evaluate
the following limit.
Solution
The two one-sided limits this time are,
The one-sided limits are the same so we get,
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There is one more limit that we need to do. However, we will need a new fact about limits
that will help us to do this.
Fact
Note that this fact should make some sense to you if we
assume that both functions are nice enough.
If both of the functions are “nice enough” to use the limit evaluation fact then we have,
The inequality is true because we know that c is somewhere between a and b and in that range we also know .
Note that we don’t really need the two functions to be nice
enough for the fact to be true, but it does provide a nice way to give a quick
“justification” for the fact.
Also, note that we said that we assumed that for all x
on [a, b] (except possibly at ).
Because limits do not care what is actually happening at we don’t really need the inequality to hold at
that specific point. We only need it to
hold around since that is what the limit is concerned
about.
We can take this fact one step farther to get the following
theorem.
Squeeze Theorem