how can we prove that a function is continuous at a certain point how can we do so there's something called the three-step continuity test and the first step is that you have to show that the function is defined at some point a so f of a has to exist it has to equal a certain value the second step is to show that the limit as x approaches a of f of x exists now how do we do that how do we show that this limit exists well you need to show that the left-sided limit the limit as x approaches a from the left side is indeed equal to the limit as x approaches a from the right side only under those conditions will the limit exist now the third step is to show that the limit as x approaches a from either side of f x is indeed equal to f of a so that's the three-step continuity test so let's go ahead and apply it with a certain example so let's say that the function f of x is equal to the square root of x plus two when x is less than two and it's equal to x squared minus two when x is between two and three and it's equal to two x plus let's say five when x is equal to or greater than three so go ahead and prove that the function is continuous or discontinuous at two and three so let's start with an x value of two so therefore we need to use these two functions so what is f of 2 to find f of 2 we need to use the second function because x is greater than or equal to two so that's going to be two squared minus two which is four minus two so that's equal to two so f of a is defined that's step one step two we need to show that the limit exists so first we need to find the value of the limit as x approaches 2 from the left side of f of x so on the left side of 2 that's when x is less than 2. so we got to use this function and so that's going to be the square root of 2 plus 2 which is the square root of 4 and that's equal to 2. so now we got to check the right side the limit as x approaches 2 from the right on the right side x is greater than 2 so we're going to use this function x squared minus 2 so that's 2 squared minus 2 which we know is 2. now because the left side and the right side are the same that means that the limit indeed exists so we can say that the limit as x approaches two from either side of f of x is equal to two now notice that these two are the same so now we can make the statement for step three that the limit as x approaches a or two of f of x is indeed equal to f of two because they both equal two so therefore the function is continuous at x equals two now let's move on to the next example and that is at an x value of three so we need to use these two functions so first let's determine if it's defined at three so f of three x is equal to three in the third part of the function so it's gonna be two times three plus five two times three is six six plus five is eleven now let's move on to step two let's find the limit as x approaches three from the left side so therefore we need to use this function three from the left is going to be less than three so it's three squared minus two three squared is nine nine minus two is seven now let's find as x approaches 3 from the right so we have to use this expression so it's going to be 2 times 3 plus 5 which is 6 plus 5 that's 11. now notice that the left side and the right side of the limit doesn't match so therefore the limit as x approaches 3 of f of x does not exist and if the limit does not exist it is not continuous at x equals 3. so therefore you could say it is discontinuous at x equal stream now because these two points do not match and we don't have a rational function this type of discontinuity is known as a jump discontinuity now let's work on some more examples here's another problem that you could try let's say that f of x is equal to 2x plus five when x is less than negative one and it's equal to x squared plus two when x is greater than negative one and then it's equal to five when x is equal to one rather negative one so there's only one x value that we need to be concerned about and that x value is negative one so go ahead and determine if it's continuous or discontinuous at negative one and if it's discontinuous determine the type of discontinuity use the three-step continuity test to do so so first we need to determine if the function is defined at negative one so what is the value of f of negative one when x is exactly negative one what is y notice that y is five when x is negative one so f of negative one is five so therefore f of a is defined so we finished with step one now step two we need to prove that the limit exists so let's find the limit as x approaches negative one from the left side so from the left side x has to be less than one i mean less than negative one so this is going to be two times negative one plus five we need to use this function so that's negative two plus five which is equal to three now let's find the limit as x approaches negative one from the right side so on the right of negative one we need to use x squared plus two because x is greater than negative one so that's going to be negative one squared plus two which is one plus two that's positive three now because these two are the same the limit exists so the limit as x approaches negative one of f of x from either side is indeed equal to three so we're finished with step two now let's focus on step three does the limit as x approaches negative one of f of x does it equal f of a notice that these two do not match they're not the same so in step three we can make the statement that the limit as x approaches negative one of f of x does not equal f of negative one so therefore step three has failed which means that it is discontinuous at negative one but the limit exists so what type of discontinuity do we have in this case if the limit exists we have this situation we have a whole but the function is not defined at the whole the limit has a y value of three but the function has a y value of five so what we have is a hole basically a removable discontinuity in the last example the jump discontinuity was a non-removable discontinuity so if step two fails if these two values are different typically it's the jump discontinuity if those two values are the same and if step three fails then usually it's going to be a whole the only time you get an infinite discontinuity is if these values equal infinity so if you don't have an infinity value it's not going to be an infinite discontinuity you