Analyzing functions for discontinuities (discontinuity example) | AP Calculus AB | Khan Academy

So we've got this function ( f(x) ) that is piecewise continuous. It's defined over several intervals. Here for ( 0 < x \leq 2 ), ( f(x) ) is ( \ln(x) ). For any ( x > 2 ), well then ( f(x) ) is going to be ( x^2 \cdot \ln(x) ).

What we want to do is we want to find the limit of ( f(x) ) as ( x ) approaches 2. What's interesting about the value 2 is that that's essentially the boundary between these two intervals. If we wanted to evaluate it at 2, we would fall into this first interval. ( f(2) ) well, 2 is less than or equal to 2 and it's greater than 0, so ( f(2) ) would be pretty straightforward. That would just be ( \ln(2) ). 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? 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 of ( f(x) ) as we approach 2 from the left, from values lower than 2. Well, this is going to be the case where we're going to be operating in this interval right over here. We're operating from values less than 2 and we're going to be approaching 2 from the left. Since this case is continuous over the interval in which we're operating, and for sure between all values greater than 0 and less than or equal to 2, this limit is going to be equal to just this clause evaluated at 2. Because it's continuous over the interval, this is just going to be ( \ln(2) ).

All right, so now let's think about the limit from the right-hand side, from values greater than 2. The limit of ( f(x) ) as ( x ) approaches 2 from the right-hand side. Well, even though 2 falls into this clause, as soon as we go anything greater than 2, we fall into this clause. So we're going to be approaching 2 essentially using this case.

Once again, this case here is continuous for all x values, not only greater than 2, actually greater than or equal to 2. For this one over here, we can make the same argument that this limit is going to be this clause evaluated at 2. Because once again if we just evaluated the function at 2, it falls under this clause. But if we're approaching from the right, well from approaching from the right those are x values greater than 2, so this clause is what's at play.

So we'll evaluate this clause at 2. Because it is continuous, this is going to be ( 2^2 \cdot \ln(2) ). So this is equal to ( 4 \cdot \ln(2) ).

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.

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, the limit does not exist.