Hi everyone! Today we’re going to talk about how to use
limit laws along with the graphs of two functions to solve a few limit problems. To complete
this problem, we’ll first identify the laws we need to use, apply them to our limit problems,
and then use information from the graphs to solve each problem.
Let’s take a look. In this particular problem, we’ve been asked
to use limit laws and the graph that we have of f of x and g of x to evaluate each of these
three limit problems. If we look at part a, we can see that we’ve
been asked to find the limit as x approaches negative two of f of x plus five times g of
x. And the first thing we need to do is identify which limit law we’re going to be using.
And what we know about limit laws is that the limit of the sum of two functions is the
same thing as the sum of the limit of each function individually. So what we can do and
all that it basically means is that we can break apart this f of x plus five times g
of x and take the limit of each one individually. We don’t have to somehow find the limit
of the two… of the combination or the sum of the two functions. So we can change that
into the limit as x approaches negative two of f of x plus the limit as x approaches negative
two of five times g of x and we haven’t changed the function at all. Limit laws tell
us that we can do this. Limit laws also tell us that the limit of a constant times a function
is the same thing as the constant times the limit of that function. Which means, again,
all that means is that we can pull the five out in front of our limit over here. So, instead
of taking the limit of five times g of x, we can take five times the limit of g of x
by itself. So, all we’re doing here is using limit laws to help us simplify this limit
problem. So now that we’ve separated these and pull the five out in front, you can see
that we’re just taking the limit as x approaches negative two of f of x and the limits as x
approaches negative two of g of x which is now totally doable. So, the limit as x approaches
negative two of f of x, we can look here at our function, f of x, and we see the point
here, negative two, right, this is… this will be the line here, x equals negative two.
So, as we approach x equals negative two of f of x we get to this point right here which
is about at y equals one, right. We have the point here, (negative two, one), so we know
that the limit as x approaches negative two of f of x is one. So we can go ahead and substitute
for this entire expression here just the value one. Then, we have plus five times the limit
as x approaches negative two of g of x. So we can see here that as we approach negative
two for g of x we get about this point right here which we’ll call (negative two, negative
one), right, because that’s about at y equals negative one there. So, we’ll substitute
for this entire expression, limit as x approaches negative two of g of x, we’ll substitute
a negative one. And now, we can just simplify this to one minus five and we get negative
four and that’s our answer for part a. Part b is the limit as x approaches one of
the product of f of x and g of x. Again, we’ll use another limit law which tells us that
the limit of the product of two functions is the same thing as the product of the limit
of the individual functions. So again, we can split this apart to say the limit as x
approaches one of f of x times the limit a s x approaches one of g of x. The thing that
we have to realize about this problem is that we can compute the limit as x approaches one
of f of x because we can see the line here, right, x equals one. So if we look at f of
x here, we have about… let’s call it (one, two), would be the point on f of x. But if
we look at g of x, we can see that we have a different left hand limit and a right hand
limit. Our left hand limit is going to be this point, right, as x approaches one, we
have the point one negative two, and our right hand limit as we come this way the point is
one negative one. So we have a different left hand limit and the right hand limit which
means that we have no general limit, that the general limit does not exist. And so,
this whole concept here of the limit as x approaches one of g of x, we can’t find.
We can find separate left and right hand limits but we can’t find a general limit for g
of x which means that this limit here does not exist and we’ll call it D N E. So, because
we can’t take the general limit of g of x, we can’t make a substitution for this
and, you know, find the limit, the general limit for f of x as x approaches one, so this
limit here does not exist. We can find separate left and right hand limits and we can do that
here. So we can say the limit as x approaches one from the negative side or from the left
of f of x times the limit as x approaches one from the negative or left hand side of
g of x and then we can go ahead and make a substitution. So, for f of x, as we approach
one from the left hand side, so we’re coming up this way and either way with f of x, it
won’t matter, but we approach the point or the line y equals two, we approach y equals
two. So we’re going to go ahead and make a substitution of two for limit as x approaches
one from the left hand side of f of x. The limit as x approaches one from the left hand
side of g of x we already know to be negative two so we plug in negative two and the answer
here we get is negative four. But, in contrast to that, if we take the limit as x approaches
one from the positive or right hand side of f of x times the limit as x approaches one
from the positive or right hand side of g of x, we’ll get, again, for f of x, we approach
the same point here, right, there is a general limit for f of x so we get two, but with g
of x, we already know, the limit here is negative one because we approach that negative one
coordinate so we plug in negative one here and we would get a value of negative two.
So this would be, we’ll call it the left hand limit, and this would be the right hand
limit which we can compute, even though we’re taking the product of these two functions
but we can’t compute the general limit. When it comes to part c here, we can see that
we have the limit as x approaches two of the quotient f of x divided by g of x and the
limit law that we want to be aware of is that the limit of a quotient is the same thing
as the quotient of limits. So we can simplify that function. Instead of taking the limit
of the quotient, we can take the limit as x approaches two of f of x all divided by
the limit as x approaches two of g of x. So that’s how we can simplify that function.
In the same way, we had trouble with part b here, we’re going to have some trouble
with part c because we never have a problem evaluating this type of a limit problem except
when the limit as x approaches two of g of x here or whatever is in the denominator is
equal to zero. So, as we can see, when x approaches two of g of x, we’re looking at this point
right here. This is x equals two on g of x and, as you can see, the value of g of x at
that point is zero. The problem with that is that regardless of what the limit is for
f of x at that point, right, of f of x, regardless of what that is, the limit as x approaches
two of g of x is zero and we can’t divide by zero. So we’re actually not able to take
this limit and we have to say that this limit does not exist because g of x is zero and
therefore this quotient here would be undefined at as x approaches two. If g of x were not
equal to zero as the… as x approached two, then we would just be able to make a substitution
for a limit as x approaches two of f of x and another one for limit as x approaches
two of g of x and we would simplify this fraction and that would be our answer. But, in this
case, because we have zero in the denominator, this limit does not exist.
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