Question Video: Understanding the Limit of a Function at a Point Mathematics • Higher Education

Given that lim_(π‘₯ β†’ 2) 𝑓(π‘₯) = 6, which of the following statements must be false? [A] 𝑓(2) = 4 [B] 𝑓(2) is undefined [C] lim_(π‘₯ β†’ 3) 𝑓(π‘₯) = 6 [D] 𝑓(2) = 6 [E] lim_(π‘₯ β†’ 2) 𝑓(π‘₯) = 4

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Video Transcript

Given that the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to six, which of the following statements must be false? Option (A) 𝑓 evaluated at two is equal to four. Option (B) 𝑓 evaluated at two is undefined. Option (C) the limit as π‘₯ approaches three of 𝑓 of π‘₯ is equal to six. Option (D) 𝑓 evaluated at two is equal to six. Or option (E) the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to four.

In this question, we’re given the limit as π‘₯ approaches two of a function 𝑓 of π‘₯. We’re told that this is equal to six. We need to use this information to determine which of five given statements is false.

So to answer this question, let’s start by recalling what we mean by the limit of a function at a point. If the values of our function 𝑓 of π‘₯ approach some finite value of 𝐿 as the values of π‘₯ approach a value of π‘Ž from either side but not necessarily when π‘₯ is equal to π‘Ž, then we say that the limit as π‘₯ approaches π‘Ž of the function 𝑓 of π‘₯ is equal to 𝐿.

In this question, we’re told that the limit as π‘₯ approaches two of a function 𝑓 of π‘₯ is equal to six. So we can substitute π‘Ž is equal to two and 𝐿 is equal to six into this definition. This then gives us the following. We know that this statement must be true, since we’re told that the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to six. And we need to use this information to determine which of the five given options cannot be true.

Let’s start with option (A), which tells us 𝑓 evaluated at two is equal to four. At first, we might be tempted to say that this cannot be true, since our values of 𝑓 of π‘₯ are approaching six when π‘₯ gets closer and closer to two. However, it’s very important to remember in our definition, we specifically say that we’re not interested in what happens when π‘₯ is equal to two. When we’re talking about limits, we’re only interested in what happens when π‘₯ gets closer and closer to the point.

To help us visualize this, let’s consider an example. Let’s consider the graph of 𝑦 is equal to π‘₯ plus four. In this graph, the 𝑦-coordinates of points on the curve represent the outputs of the function. We want to use this to determine the limit as π‘₯ approaches two of π‘₯ plus four. Before we do this, we know that the point with coordinates two, six lies on this line, since if we substitute π‘₯ is equal to two into the equation, we get that 𝑦 is equal to six.

Let’s now see what happens to the output values of this function as our values of π‘₯ approach two from either side. First, as our values of π‘₯ approach two from above, we can see that our output values, the 𝑦-coordinates, are approaching six from above. Similarly, we can see if our input values of π‘₯ approach two from the left, then our output values are approaching six from below. Therefore, since the outputs of our function are approaching six as π‘₯ approaches two from either side, we’ve shown the limit as π‘₯ approaches two of π‘₯ plus four is equal to six.

And this is a useful result. If we call our function 𝑓 of π‘₯ π‘₯ plus four, then we’ve shown the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to six, which is 𝑓 evaluated at two. And this is an example which shows that option (D) can be true, since we have the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to six, which is also 𝑓 evaluated at two. But we can also directly manipulate this example to show that option (A) can also be true.

We want to change our function so that 𝑓 evaluated at two is equal to four. And the easiest way to do this is just to change the single output value. We’ll set our function 𝑓 of π‘₯ to be equal to π‘₯ plus four everywhere except when π‘₯ is equal to two. In this case, we’ll output four.

We can then notice something interesting. First, the graph of this function will be incredibly similar to the graph of the straight line 𝑦 is equal to π‘₯ plus four, the only difference being the single output value change. We can then use the exact same method we did before to determine the limit as π‘₯ approaches two of 𝑓 of π‘₯. Remember, when we’re taking our limit at two, we’re not interested in what happens to our function when π‘₯ is equal to two, only what happens when π‘₯ gets closer and closer to two.

So once again, as our input values of π‘₯ approach two from the right, our outputs will approach six. And as our values of π‘₯ approach two from the left, our outputs will also approach six. So the limit as π‘₯ approaches two of this new function 𝑓 of π‘₯ is equal to six. However, 𝑓 evaluated at two is equal to four. Therefore, this is an example of a function 𝑓 of π‘₯ where option (A) holds true. The limit of this function is six when π‘₯ approaches two; however, 𝑓 evaluated at two is equal to four. And it’s worth noting we could have chosen any output value for this point.

We can now move on to option (B). However, we can see that we can also manipulate this example to show that option (B) can also hold true. In exactly the same way, the easiest way to make 𝑓 evaluated at two undefined will just be to remove the output value. This time, we’ll set our function 𝑓 of π‘₯ equal to the linear function π‘₯ plus four for all values of π‘₯ not equal to two. And we’ll leave our function undefined when π‘₯ is equal to two. And then the graph of this function is the straight line 𝑦 is equal to π‘₯ plus four with a single point removed at two, six.

Once again, we have not changed any of the output values of our function around this point. We’ve only changed the value of the function when π‘₯ is equal to two, which doesn’t change its limit. So the limit as π‘₯ approaches two of this function is six. And in this function, 𝑓 of two is not defined. Therefore, this is an example where option (B) holds true. The limit as π‘₯ approaches two of this function is six. However, this function is not defined when π‘₯ is equal to two.

Let’s now move on to option (C). We could construct another entirely different example to make this hold true. However, we’re just going to manipulate this example one more time to make option (C) hold true. Since we want the limit as π‘₯ approaches three of our function 𝑓 of π‘₯ to be equal to six, we’re going to construct a function which is the straight line 𝑦 is equal to π‘₯ plus four combined with the horizontal line 𝑦 is equal to six. And for due diligence, let’s write this out as a piecewise function. 𝑓 of π‘₯ will be equal to π‘₯ plus four when π‘₯ is less than or equal to two, and 𝑓 of π‘₯ will be equal to the constant value of six when π‘₯ is greater than two.

We can now determine both the limit as π‘₯ approaches two of 𝑓 of π‘₯ and the limit as π‘₯ approaches three of 𝑓 of π‘₯. Let’s start with the limit as π‘₯ approaches three of 𝑓 of π‘₯. We need to see what happens to the output values of this function as π‘₯ approaches three from either side. Well, as π‘₯ approaches three from the right, all of our outputs are equal to six. Similarly, as π‘₯ approaches three from the left, all of the output values are equal to six. So the limit as π‘₯ approaches three of 𝑓 of π‘₯ is equal to six.

Let’s now use the diagram to determine the limit as π‘₯ approaches two of this function. First, as our values of π‘₯ approach two from the right, our output values remain constant at six. Next, as our values of π‘₯ approach two from the left, we can see that they’re also approaching six. Therefore, the limit as π‘₯ approaches two of 𝑓 of π‘₯ is also equal to six. Therefore, this is a function where the limit as π‘₯ approaches two of 𝑓 of π‘₯ and the limit as π‘₯ approaches three of 𝑓 of π‘₯ are equal to six.

It is worth noting, however, this is not the easiest example we could have chosen. We could have just chosen our function 𝑓 of π‘₯ to be a constant value of six. Then, for any real value π‘Ž, the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ will always be equal to six, since the function always outputs six.

This leaves us with option (E). We need to show that this is false. To show that this value is false, let’s consider what happens when 𝐿 is four in the definition of this limit. This would then tell us that the values of 𝑓 of π‘₯ approach four as π‘₯ approaches two from either side but not necessarily when π‘₯ is equal to two. However, we also need this limit to be equal to six. This means we need the values of 𝑓 of π‘₯ to approach four and approach six as our values of π‘₯ approach two from either side. The outputs can’t get closer and closer to both four and six; these are different numbers. And this is true in general. The limit as π‘₯ approaches π‘Ž of any function 𝑓 of π‘₯ can’t have two distinct values. It can only have one value.

Therefore, we were able to show if the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to six, then of the five given options, only option (E), the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to four, is false.

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