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

So we have ( g(x) ) being defined as the log of ( 3x ) when ( 0 < x < 3 ) and ( 4 - x ) times the log of ( 9 ) when ( x \geq 3 ).

So based on this definition of ( g(x) ), we want to find the limit of ( g(x) ) as ( x ) approaches ( 3 ). Once again, this ( 3 ) is right at the interface between these two clauses or these two cases.

We go to this first case when ( x ) is between ( 0 ) and ( 3 ), when it's greater than ( 0 ) and less than ( 3 ), and then at ( 3 ) we hit this case. In order to find the limit, we want to find the limit from the left-hand side, which will have us dealing with this situation. Because if we're less than ( 3 ), we're in this clause, and we also want to find the limit from the right-hand side, which would put us in this clause right over here.

Then, if both of those limits exist and if they are the same, then that is going to be the limit of this. So let's do that.

Let me first go from the left-hand side, so the limit as ( x ) approaches ( 3 ) from values less than ( 3 ). So we're going to approach from the left of ( g(x) ). Well, this is equivalent to saying this is the limit as ( x ) approaches ( 3 ) from the negative side when ( x ) is less than ( 3 ).

Which is what's happening here; we’re approaching ( 3 ) from the left. We're in this clause right over here, so we're going to be operating right over there. That is what ( g(x) ) is when we are less than ( 3 ), so we have ( \log(3x) ).

Since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all ( x > 0 ). Well, we can just substitute ( 3 ) in here to see what it would be approaching. So this would be equal to ( \log(3 \times 3) ) or ( \log(9) ).

And once again, when people just write ( \log ) here without writing the base, it's implied that we're dealing that it is ( 10 ) right over here. So this is ( \log_{10} ). That's just a good thing to know that sometimes gets missed a little bit.

All right, now let's think about the other case. Let's think about the situation where we are approaching ( 3 ) from the right-hand side, from values greater than ( 3 ). Well, we are now going to be in this scenario right over there.

So this is going to be equal to the limit as ( x ) approaches ( 3 ) from the positive direction, from the right-hand side of ( g(x) ) in this clause when we are greater than ( 3 ), so ( 4 - x ) times ( \log(9) ).

And this looks like some type of a logarithm expression at first until you realize that ( \log(9) ) is just a constant. ( \log_{10}(9) ) is going to be some number close to ( 1 ). This expression would actually define a line for ( x \geq 3 ).

( g(x) ) is just a line, even though it looks a little bit complicated, and so this is actually defined for all real numbers. It's continuous for any ( x ) that you put into it, so to find this limit, we think about what this expression is approaching as we approach ( 3 ) from the positive direction. Well, we can just evaluate it at ( 3 ).

So it's going to be ( 4 - 3 ) times ( \log(9) ), well that's just ( 1 ), so that's equal to ( \log_{10}(9) ).

So the limit from the left equals the limit from the right; they're both ( \log(9) ). So the answer here is ( \log(9) ), and we are done.