Question Video: Comparing Continuity at a Point to the Function Evaluated at the Point Mathematics • Higher Education

If 𝑓(6) = βˆ’6, what can we say about lim_(π‘₯ β†’ 6) 𝑓(π‘₯)?

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Video Transcript

If 𝑓 evaluated at six is equal to negative six, what can we say about the limit as π‘₯ approaches six of 𝑓 of π‘₯? Option (A) the limit as π‘₯ approaches six of 𝑓 of π‘₯ is equal to negative six. Option (B) the limit as π‘₯ approaches six of 𝑓 of π‘₯ is not equal to negative six. Option (C) the limit as π‘₯ approaches six of 𝑓 of π‘₯ is equal to zero. Option (D) we cannot draw any conclusions about the limit as π‘₯ approaches six of 𝑓 of π‘₯. Or option (E) the limit as π‘₯ approaches six of 𝑓 of π‘₯ is equal to negative one.

In this question, we’re given a function 𝑓 of π‘₯. And we’re only told one piece of information about this function: 𝑓 evaluated at six is negative six. We need to determine whether we can use this information to determine anything about the limit as π‘₯ approaches six of 𝑓 of π‘₯. We’re given five possible options.

To answer this question, let’s start by recalling what we mean by the limit of a function at a point. We recall if the values of our function 𝑓 of π‘₯ approach some finite value of 𝐿 as the values of π‘₯ approach 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 our case, we’re interested in the limit of a function 𝑓 of π‘₯ as π‘₯ approaches six. So let’s set our value of π‘Ž equal to six in this definition.

This gives us the following. And we can immediately notice something from the definition. We’re not interested in the function’s output when our value of π‘₯ is equal to six. We’re only interested in what happens to the output values of our function as the input values of π‘₯ approach six. They get closer and closer to six, but they’re never equal to six. And this is actually enough to answer this question. We can’t draw any conclusions about the limit of 𝑓 of π‘₯ as π‘₯ approaches six by being told the value of 𝑓 evaluated at six because in the definition we’re not interested in the value of our function at the limit point.

So we could stop here and just say that the answer is option (D). However, it can be useful to explicitly show that options (A), (B), (C), and (E) are not correct. And there’s a few different ways of doing this.

We’re going to approach this problem graphically. Let’s start by thinking of a function 𝑓 of π‘₯, where 𝑓 evaluated at six is equal to negative six. And we can easily determine the limit as π‘₯ approaches six of 𝑓 of π‘₯. And arguably, the easiest choice of function will be a constant function. 𝑓 of π‘₯ is just the constant value of negative six. And we know the sketch of this function will be a horizontal line 𝑦 is equal to negative six.

We can now determine the limit of this function from its graph. As our values of π‘₯ approach six from the left, we can see the outputs of our function are a constant value of negative six. Similarly, as the values of π‘₯ approach six from the right, we can see the outputs are a constant value of negative six. Therefore, if 𝑓 of π‘₯ is a constant value of negative six, we’ve shown the limit of 𝑓 of π‘₯ as π‘₯ approaches six is equal to negative six.

And this shows us three things. First, the limit as π‘₯ approaches six of 𝑓 of π‘₯ can be equal to negative six, so option (B) is incorrect. Second, the limit as π‘₯ approaches six of 𝑓 of π‘₯ does not have to be equal to zero, so option (C) is not correct. Finally, the limit as π‘₯ approaches six of 𝑓 of π‘₯ does not have to be equal to negative one, so option (E) is also not correct.

However, we’ve not shown that option (A) is incorrect because this agrees with our current function 𝑓 of π‘₯. So let’s clear some space and show that this does not have to be true. We want to construct a function 𝑓 of π‘₯ where 𝑓 evaluated at six is equal to negative six but the limit as π‘₯ approaches six of 𝑓 of π‘₯ is not equal to negative six.

First, we’re only given one piece of information about 𝑓 of π‘₯; 𝑓 evaluated at six needs to be equal to negative six. This means when we draw the graph of our function, we’re only told one point where the function passes through: six, negative six. The rest of the graph of this function could be anything we want. For example, we could just say that our function outputs six for all other values of π‘₯. We can even write this as a piecewise function. 𝑓 of π‘₯ is equal to six if π‘₯ is not equal to six, and 𝑓 of π‘₯ is equal to negative six if π‘₯ is equal to six.

And now we can notice something interesting. As our values of π‘₯ approach six from the right, we can see the output values of the function remain at a constant value of six. Similarly, as our values of π‘₯ approach six from the left, we can see the output values of the function still remain at a constant value of six. Therefore, the limit of this function 𝑓 of π‘₯ as π‘₯ approaches six is equal to six. And this does not agree with option (A), which says that this limit would need to be equal to negative six.

And it’s worth noting we can adapt this to show that the limit as π‘₯ approaches six of 𝑓 of π‘₯ can be equal to any value. All we would need to do is change the value of our horizontal line. If this was a value of 𝑐, then the outputs of our function won’t be at a constant value of 𝑐. So the limit as π‘₯ approaches six of 𝑓 of π‘₯ would be equal to 𝑐. This, then, shows that there’s no relation between 𝑓 evaluated at six and the limit as π‘₯ approaches six of 𝑓 of π‘₯ in a general case, which is another way of showing that we cannot draw any conclusions about the limit as π‘₯ approaches six of 𝑓 of π‘₯ from 𝑓 evaluated at six, which was option (D).

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