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# Worked example: point where a function isn't continuous

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.A (LO)
,
LIM‑2.A.2 (EK)

## Video transcript

so we've got this function f of X that is piecewise continuous is defined over several intervals here for X being or for 0 less than X being less than or equal to 2 f of X is natural log of X for any X is larger than 2 well then f of X is going to be x squared times the natural log of X and what we want to do is we want to find the limit of f of X as X approaches 2 and what's interesting about the value two is that that's at the that's the essentially the boundary between these two intervals if we wanted to evaluate it at two we would fall into this first interval F of 2 well 2 is less than or equal to 2 it's and it's greater than 0 so f of 2 f of 2 would be pretty straight forward that would just be natural log of 2 but that's not necessarily what the limit is going to be before defiant 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 let's do that let's first think about the limit the limit of f of 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 and so will fall under this clause and so since this this clause or case is continuous over the interval in which we're operating and for sure' between or for all values greater than 0 and less than or equal to 2 this limit is going to be equal to just this clause evaluated it to because it's continuous over the interval so this is just going to be the natural log of 2 all right so now let's think about the limit from the right hand side from values greater than 2 so the limit the limit of f of X as X approaches 2 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 going to be approaching two essentially using this case and once again this case here is is is continuous for all X values not only greater than to actually you know greater than or equal to two and so for 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 were just evaluated the function at two it falls into this clause but if we're approaching from the right well if we're 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 so because it is continuous so this is going to be 2 squared times the natural log of 2 and so this is equal to 4 times the natural log of 2 4 times the natural log of 2 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 that jump discontinuity this limit would not exist because the left handed limit and the right hand limit go to two different values so limit so this does not exist
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