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Current time:0:00Total duration:4:03

AP.CALC:

LIM‑2 (EU)

, LIM‑2.A (LO)

, LIM‑2.A.2 (EK)

- [Voiceover] So we've
got this function f of x that is a piecewise continuous. It's defined over several intervals here for x being, or for zero less than x, and being less than or equal to two. F of x is natural log of x. For any x's larger than two, well then, f of x is going to be x squared times the natural log of x. And what we wanna do is we wanna find the limit of f of x as x approaches two. Now, what's interesting
about the value two is that that's essentially
the boundary between these two intervals. If we wanted to evaluate it at two, we would fall into this first interval. F of two, well, two is
less than or equal to two, and it's greater than zero. So f of two would be
pretty straightforward. That would just be natural log of two. But that's not necessarily
what the limit is going to be. To figure out what the
limit is going to be, we should think about, well, what's the limit as we
approach from the left, what's the limit as we
approach from the right, and do those exist, and if they do exist, are they the same thing, and if they are the same thing, well then, we have a well-defined limit. So let's do that. Let's first think about the limit. The limit of f of x as we approach two from the left, from values lower than two. Well, this is gonna be the case where we're gonna be
operating in this interval right over here. We're operating from values less than two, and we're going to be approaching two from the left. And so we'll fall under this clause, and so, since this clause or case is continuous over the interval in which we're operating, and for sure between, or for
all values greater than zero and less than or equal to two, this limit is going to be equal to just this clause evaluated at two, because it's continuous over the interval. So this is just going to
be the natural log of two. All right, so now let's
think about the limit from the right hand side, from values greater than two. So the limit of f of x as x approaches two from
the right hand side. Well, even though two
falls into this clause, as soon as we go anything
greater than two, we fall in this clause. So we're gonna be approaching two essentially using this case. And once again, this case here is continuous for all x values not only greater than two, actually, you know, greater
than or equal to two. And so, for this one over here, we can make the same
argument that this limit is going to be this
clause evaluated at two, because once again, if we just evaluated the function at two, it
falls under this clause, but if we're approaching from the right, if we're approaching from the right, those are x values greater than two, so this clause is what's at play. So we'll evaluate this clause at two. So, because it is continuous. So this is going to be two squared times the natural log of two. And so this is equal to four times the natural log of two. Four times the natural log of two. So, the right hand limit does exist. The left hand limit does exist. But the thing that might jump out at you is that these are two different values. We approach a different
value from the left as we do from the right. If you were to graph this, you would see a jump in the actual graph. You would see a discontinuity
occurring there. And so for this one in particular, you have that jump discontinuity, this limit would not exist because the left hand limit
and the right hand limit go to two different values. So this does not exist.