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 ofx exists now how do we do that how do we show that this limit exists well you need to show that the left side 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 exists now the third step is to show that the limit as X approaches a from either side of f ofx 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 ofx is equal to the sare < TK of x + 2 when X is less than 2 and it's equal to x^2 - 2 when X is between 2 and 3 and it's equal to 2x plus let's say 5 when X is equal to or greater than three so go ahead and prove that the function is Contin ous or discontinuous at 2 and 3 so let's start with an x value of two so therefore we need to use these two functions so what is f of two to find F of two we need to use the second function because X is greater than or equal to 2 so that's going to be 2^ 2 - 2 which is 4 - 2 so that's equal to 2 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 two from the left side of f ofx so on the left side of two that's when X is less than two so we got to use this function and so that's going to be the square < TK of 2 + 2 which is the square < TK of 4 and that's equal to two so now we got to check the right side the limit as X approaches two from the on the right on the right side x is greater than two so we're going to use this function x^2 - 2 so that's 2^ 2us 2 which we know is two 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 ofx is equal to 2 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 2 of f ofx is indeed equal to F of two because they both equal two so therefore the function is continuous at x = 2 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 uh functions so first let's determine if it's defined at three so F of 3 x is equal to 3 in the third part of the function so it's going to be 2 * 3 + 5 2 * 3 is 6 6 + 5 is 11 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 3 from the left is going to be less than 3 so it's 3^ 2 - 2 3^ 2 is 9 9 - 2 is 7 now let's find as X approaches three from the right so we have to use this expression so it's going to be 2 * 3 + 5 which is 6 + 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 ofx does not exist and if the limit does not exist it is not continuous at xal 3 so therefore you could say it is discontinuous at xal 3 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 ofx is equal to 2x + 5 when X is less than1 and is equal to x^2 + 2 when X is greater than1 and then it's equal to 5 when X is equal to 1 rather negative 1 so there's only one x value that we need to be concerned about and that x value is 1 so go ahead and determine if it's continuous or discontinuous at1 and if it's discontinous 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 at1 so what is the value of f of1 when X is exactly ne1 what is y notice that Y is 5 when X is 1 so f of1 is is five so therefore F of a is defined so we finished with uh step one now step two we need to prove that the limit exists so let's find the limit as X approaches -1 from the left side so from the left side x has to be less than one I mean less than negative 1 so this is going to be 2 * 1 + 5 we need to use this function so that's -2 + 5 which is equal to 3 now let's find the limit as X approaches ne1 from the right side so on the right of negative 1 we need to use x^2 + 2 because X is greater than 1 so that's going to be -12 + 2 which is 1 + 2 that's posi 3 now because these two are the same the limit exists so the limit as X approaches -1 of f ofx from either side is indeed equal to three so we finished with step two now let's focus on step three does the limit as X approaches Nega 1 of f ofx 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 -1 of f ofx does not equal F of1 so therefore step three has failed which means that it is discontinuous at netive 1 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 hole but the function is not defined at the hole the limit has a yvalue of three but the function has a yvalue 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 hole 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 e