Question Video: Determining the Limit of a Function from Its Graph at a Point of Jump Discontinuity

Determine the limit of the function as π‘₯ ⟢ 2.

03:10

Video Transcript

Determine the limit of the function as π‘₯ approaches two.

We’re given a graph of a function 𝑓 of π‘₯. We need to determine the limit of this function as π‘₯ is approaching two. If we were to try and do this directly from our graph, we would arrive at a problem. If we take values of π‘₯ close to two but less than two, we can see that our outputs are less than zero. However, if we take our values of π‘₯ close to two but greater than two, we can see that our outputs are bigger than three. And we know that the outputs of this function can’t be approaching both zero and three.

So to answer this question, we’re going to need to recall the relationship between left-hand and right-hand limits and the regular limits. We recall if the limit as π‘₯ approaches two from the left of 𝑓 of π‘₯ is equal to some finite value of 𝐿 and the limit as π‘₯ approaches two from the right of 𝑓 of π‘₯ is also equal to 𝐿, then we must have the limit as π‘₯ approaches two of 𝑓 of π‘₯ is also equal to 𝐿. In fact, this works in reverse. If we know the limit as π‘₯ approaches two of 𝑓 of π‘₯ is equal to some finite value of 𝐿, then both the limit as π‘₯ approaches two from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches two from the right of 𝑓 of π‘₯ must also be equal to 𝐿.

But we need to remember what happens if these conditions are not met. If the left-hand limit and the right-hand limit are not equal, then we say the limit as π‘₯ approaches two of 𝑓 of π‘₯ does not exist. So to answer this question, we can look at the left- and right-hand limit individually. Let’s start with the limit as π‘₯ approaches two from the left of 𝑓 of π‘₯. Our values of π‘₯ are approaching two from the left, so π‘₯ will be less than two. We want to see what happens to our outputs of 𝑓 of π‘₯ as π‘₯ gets closer and closer to two from the left.

From our graph, we can see that 𝑓 of negative three is equal to negative five. Moving closer to two, we can see when π‘₯ is equal to negative one, our output 𝑓 of negative one is equal to negative three. Moving closer still, when π‘₯ is equal to one, we can see our function outputs negative one. And in fact, from this sketch, we can see as π‘₯ approaches two from the left, our output values are getting closer and closer to zero. So as π‘₯ approached to two from the left, our outputs 𝑓 of π‘₯ go closer and closer to zero. This is the same as saying the limit as π‘₯ approaches two from the left of 𝑓 of π‘₯ is equal to zero.

We can do exactly the same as π‘₯ approaches two from the right. Let’s start with π‘₯ is equal to seven. We can see from our graph that 𝑓 of seven is equal to eight. From our graph, we can also see that 𝑓 of five is equal to six. When π‘₯ is equal to three, we can see our function outputs four. And as we get closer and closer to two from the right, we can see that our outputs are getting closer and closer to three.

So from our graph, we can conclude the limit as π‘₯ approaches two from the right of 𝑓 of π‘₯ is equal to three. But this means we’ve shown that our limit as π‘₯ approached to two from the left was not equal to our limit as π‘₯ approached to two from the right. So this means that the limit as π‘₯ approaches two of 𝑓 of π‘₯ must not exist.

Therefore, by looking at the graph of this function, we were able to determine the limit as π‘₯ approached to two from the left of 𝑓 of π‘₯ was equal to zero and the limit as π‘₯ approached two from the right of 𝑓 of π‘₯ was equal to three. And then, because these two values were not equal, we were able to conclude that the limit as π‘₯ approaches two of 𝑓 of π‘₯ does not exist.

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