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  • - [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.

- [Voiceover] So we've got this function f of x

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