Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists | Nagwa Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists | Nagwa

Question Video: Finding the Limit of a Function from Its Graph at a Point of Removable Discontinuity If the Limit Exists Mathematics

Determine lim_(π‘₯ β†’ 4) 𝑓(π‘₯), if it exists.

02:27

Video Transcript

Determine the limit as π‘₯ tends to four of 𝑓 of π‘₯, if it exists.

We have a graph of 𝑓 of π‘₯ and we want to use it to see what happens as π‘₯ tends to four. We can see from this point here that 𝑓 of four is equal to one. But this doesn’t help us because we’re looking for the limit as π‘₯ tends to four. And so, we look at the value of the function 𝑓 when π‘₯ is near but not equal to four. When π‘₯ is two, 𝑓 of π‘₯ is one. When π‘₯ is closer to four, π‘₯ is three, 𝑓 of π‘₯ is zero. And we can get closer still, when π‘₯ is 3.5, 𝑓 of π‘₯ is negative 0.5. We can see that as π‘₯ gets closer and closer to four from below, 𝑓 of π‘₯ gets closer and closer to negative one.

And we can write down this fact like this: the limit as π‘₯ tends to four from below of 𝑓 of π‘₯ is equal to negative one. So we’ve looked at what happens when π‘₯ approaches four from below, that is π‘₯ is less than four but getting closer and closer to four.

Now we need to look at what happens when π‘₯ approaches four from above. 𝑓 of six is one. As π‘₯ gets closer to four, 𝑓 of π‘₯ gets closer to negative one. So 𝑓 of five is closer to negative one than 𝑓 of six was; 𝑓 of five is zero. And this trend continues as π‘₯ approaches four from above, that is π‘₯ is greater than four but getting closer and closer to four. We can see that 𝑓 of π‘₯ gets closer and closer to negative one. And so the limit as π‘₯ tends to four from above of 𝑓 of π‘₯ is also equal to negative one.

The limits from above and from below are equal. And so we can say that the limit as π‘₯ tends to four, period, of 𝑓 of π‘₯ exists and is equal to negative one. Notice however that the limit as π‘₯ tends to four of 𝑓 of π‘₯, which we found was negative one, is not equal to the value of 𝑓 of four which, according to the graph, is one. So the limit of a function at a point can exist but not be equal to the value of the function at that point.

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