Video: Evaluating the Limit of a Function from Its Graph

The following figure is the graph of the function 𝑓, where 𝑓(π‘₯) = ((4(π‘₯ βˆ’ 5)√(π‘₯ βˆ’ 1)) βˆ’ ((π‘₯ βˆ’ 1)(π‘₯ βˆ’ 5)))/(π‘₯ βˆ’ 5). What is the value of 𝑓(5)? What does the graph suggest about the value of lim_(π‘₯ β†’ 5) 𝑓(π‘₯)?

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Video Transcript

The following figure is the graph of the function 𝑓, where 𝑓 of π‘₯ is equal to four times π‘₯ minus five multiplied by the square root of π‘₯ minus one minus π‘₯ minus one times π‘₯ minus five all divided by π‘₯ minus five. What is the value of 𝑓 evaluated at five? What does the graph suggest about the value of the limit as π‘₯ approaches five of 𝑓 of π‘₯?

In this question, we’re given the graph of a function 𝑓 of π‘₯ and we’re actually given an explicit formula for our function 𝑓 of π‘₯. We need to use both of these pieces of information to determine two things. First, we need to determine the value of 𝑓 evaluated at five. And one way of doing this because we’re given an explicit definition for our function 𝑓 of π‘₯ is just to substitute π‘₯ is equal to five into our function 𝑓 of π‘₯.

Substituting π‘₯ is equal to five into 𝑓 of π‘₯, we get 𝑓 of five is four times five minus five multiplied by the square root of five minus one minus five minus one times five minus five all divided by five minus five. And if we evaluate this expression, we see both terms in our numerator have a factor of zero and our denominator evaluates to give us zero. In other words, 𝑓 evaluated at five evaluates to give us zero divided by zero. And we know we can’t divide zero by zero. So 𝑓 evaluated at five must be undefined. So this is one way of showing that 𝑓 evaluated at five is undefined.

However, we can do this directly from the graph of 𝑓 of π‘₯ as well. Since this is a graph of the curve 𝑦 is equal to 𝑓 of π‘₯, our values of π‘₯ will be our inputs to our function and our values of 𝑦 will be the output of our function. We want to know what happens when our input value of π‘₯ is equal to five. So we can do this by looking at our graph. And we can see something very interesting about our curve when our value of π‘₯ is equal to five. We can see we have a hollow circle on our diagram. And whenever we see a hollow circle like this in a curve, we know that this means our function is undefined at this point. Therefore, our function 𝑓 of π‘₯ is undefined when our input value of π‘₯ is equal to five.

Now, the second part of this question wants us to answer, what does the graph suggest about the value of the limit as π‘₯ approaches five of 𝑓 of π‘₯? And to start answering this question, we’re going to first need to recall what we mean by the limit as π‘₯ approaches some value of a function. We recall we say that the limit as π‘₯ approaches π‘Ž of some function 𝑓 of π‘₯ is equal to some constant 𝐿 if the values of our function 𝑓 of π‘₯ approach 𝐿 as π‘₯ is approaching our value of π‘Ž from both sides.

In our case, we want to know the limit as π‘₯ approaches five of our function 𝑓 of π‘₯. So we want our value of π‘Ž equal to five. In fact, we can just update our definition with the value of π‘Ž set to be five. Let’s take a closer look at our definition. We want to know what happens to the outputs of our function 𝑓 of π‘₯ as our values of π‘₯ are approaching five from both sides. And in fact, we can do this directly from our graph.

Let’s start by seeing what happens to our outputs of 𝑓 of π‘₯ as π‘₯ approaches five from above. And it’s also worth pointing out this is usually called from the right, because when our values of π‘₯ are bigger than five, this means on our number line or on our axis our values of π‘₯ will be to the right of five. We want to see what happens to our outputs of π‘₯ as π‘₯ gets closer to five from the right.

So we’re going to need to start with some value of π‘₯. Let’s start with π‘₯ is equal to eight. We can approximate the value of 𝑓 evaluated at eight by using our graph. We go up vertically from our input π‘₯ is equal to eight up to our curve. And then we go horizontally across to find the 𝑦-coordinate. And we can see in this case our output is approximately equal to 3.6.

And it’s also worth pointing out, in this case, we could find 𝑓 evaluated at eight directly. All we need to do is substitute π‘₯ is equal to eight into our definition of the function 𝑓 of π‘₯. However, we’ll see this is not necessary to answer our question. Remember, we want to know what happens as our values of π‘₯ are approaching five from both directions. So our values of π‘₯ need to get closer and closer to five. So let’s do the exact same thing with a value of π‘₯ even closer to five. Let’s try π‘₯ is equal to six.

Doing exactly the same thing we did before, we can approximate 𝑓 evaluated at six. This time, we can see it’s very, very close to four. It’s approximately 3.9. And we can keep doing the same thing. And then we can notice something interesting. As our values of π‘₯ are getting closer and closer to five from the right, our outputs of 𝑓 of π‘₯ are getting closer and closer to four. This means we’re going to want to try setting our value of 𝐿 equal to four. But we’re not done yet.

Remember, we need to know what happens to our outputs of 𝑓 of π‘₯ as π‘₯ approaches five from both sides. So we need to know what happens as our values of π‘₯ approach five from the left. And we can do this in exactly the same way. We need to start with some value of π‘₯ less than five. Let’s start with π‘₯ is equal to two.

From our graph, we see when π‘₯ is equal to two, our function 𝑓 of π‘₯ is approximately equal to three. In fact, if we substituted π‘₯ is equal to two into our function, we would see that it’s exactly equal to three. But we don’t need to know this. We just want to know what happens to our outputs as π‘₯ gets closer and closer to five. So let’s try an input of π‘₯ even closer to five. Let’s try π‘₯ is equal to four.

Doing exactly the same thing we did before, we can approximate 𝑓 evaluated at four from our graph. This time, we see it’s also approximately equal to 3.9. And we can do exactly the same thing, choosing values of π‘₯ closer and closer to five from the left. And we’ll see something interesting. As our values of π‘₯ get closer and closer to five from the left, we see our outputs of 𝑓 of π‘₯ are getting closer and closer to four. So as π‘₯ approach five from the left, our outputs got closer to four. And as π‘₯ approach five from the right, our outputs also got closer to four. But this is exactly what we mean when we say the limit as π‘₯ approaches five of 𝑓 of π‘₯ is equal to four.

Therefore, we’ve shown that the graph suggests that the limit as π‘₯ approaches five of 𝑓 of π‘₯ is equal to four. And there’s something interesting we can see about this question. In the first part of this question, we showed that 𝑓 evaluated at five is undefined. However, in the second part of this question, we showed that the limit as π‘₯ approaches five of 𝑓 of π‘₯ is equal to four. In other words, we’ve shown even though the function is not defined when π‘₯ is equal to five, its limit as π‘₯ approaches five exists and is equal to four. And there’s an important reason for this.

In our definition of the limit, we wonder what happens to our outputs as π‘₯ approaches five. This means we want to know what happens as our values of π‘₯ are getting closer and closer to five. π‘₯ is never equal to five. And this is why our function not being defined when π‘₯ is equal to five won’t change the value of its limit as π‘₯ approaches five.

Therefore, we were able to show, given a function 𝑓 of π‘₯ and a graph of this function 𝑓 of π‘₯, the function will be undefined when π‘₯ is equal to five. And the graph suggests the limit as π‘₯ approaches five of 𝑓 of π‘₯ will be equal to four.

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