Video: Finding the One-Sided Limit of a Function from Its Graph at a Point If the Limit Exists

Find lim_(π‘₯ β†’ βˆ’3⁻) 𝑓(π‘₯).

03:02

Video Transcript

Find the limit as π‘₯ tends to negative three from below of 𝑓 of π‘₯.

We have a graph of the function 𝑓 of π‘₯. And we need to use it to find the limit as π‘₯ tends to negative three from below of this function. Let’s write out this limit again. The value that π‘₯ tends to is negative three. So we’re asking what value does 𝑓 of π‘₯ gets closer and closer to as π‘₯ gets closer and closer to negative three. Okay, now how about this superscript minus sign next to the negative three? What does this mean?

Looking at a number line, we can see that there are two ways that π‘₯ can get closer and closer to negative three. π‘₯ can approach negative three from the left, passing negative five then negative four then negative 3.5 then negative 3.1 and so on. In this case, π‘₯ is always less than negative three as it approaches negative three. Alternatively, π‘₯ can approach negative three from the right. In this case, π‘₯ is always greater than negative three as it tends to negative three. Taking the value zero then negative one then negative two then negative 2.9 then negative 2.999 and so on always been greater than negative three but getting closer and closer to it.

The superscript minus sign in the limit tells us that we only care about the case when π‘₯ is less than negative three, as it approaches negative three. Let’s have a look at our graph then. π‘₯ is approaching negative three from the left with π‘₯ less than negative three. The question is what value does 𝑓 of π‘₯ approach as π‘₯ changes in this way. For example, we can read off the value of 𝑓 of negative five. This is negative two. In a similar way, we can see that 𝑓 of negative four is one. 𝑓 of negative 3.5 looks to be just below two. And as π‘₯ gets even closer to negative three from this direction, 𝑓 of π‘₯ looks like it’s getting closer to two. So this is our answer. The limit as π‘₯ tends to negative three from below of 𝑓 of π‘₯ is two. The value that 𝑓 of π‘₯ gets closer and closer to as π‘₯ gets closer and closer to negative three, importantly with π‘₯ less than negative three, is two.

If instead we had to find the limit as π‘₯ tends to negative three with this superscript plus sign of 𝑓 of π‘₯, then we would be interested in what value 𝑓 of π‘₯ gets closer and closer to as π‘₯ gets closer and closer to negative three, with π‘₯ greater than negative three. Again, we can use our graph for this. As π‘₯ gets closer and closer to negative three with π‘₯ greater than negative three, 𝑓 of π‘₯ is getting closer and closer to zero. And so the value of this limit is zero.

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