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Video transcript

problem number six let F be defined by f of X is equal to and we have two cases when X is less than or equal to 0 F is 1 minus 2 sine of X when x is greater than 0 F is e to the negative 4x show that F is continuous at x equals 0 so for something to be continuous at x equals 0 let's think about what has to happen so if I have a function here so that is my x-axis and let's say this is my this is my y-axis and we care about what happens at x equals 0 so x equals 0 there and let's say that this is our function let's say that this is our function so maybe our function may be look something like this I don't know what well this one in particular probably looks something like this this one in particular this one in particular might look who knows and then it might look something like that and the particulars aren't that important we just have to think about what they're asking in order for it to be continuous here the limit as we approach as we approach from the left the limit as we approach from the left so the limit as we approach 0 from the left should be equal to the value of the function at 0 so this should so the limit of f of X should be equal to f of 0 which should be equal to the limit as we approach from the right which should be equal to the limit as we approach 0 from the right so that should be equal to the limit as X approaches 0 from the right of f of X and the reason why this matters is if this if this wasn't the case if F of 0 wasn't the same as the limit then we might have a gap right over there so you could have a you could have limits so you could have a situation like this where you have a gap right over there and then it looks something like that so the two limits from the left and the right both exist and the limit at that point would exist but if the function itself does not equal that value there if it equals something else then the function would not be continuous so that's why the limit has to be equal to the value of the function in order for it to be continuous so let's think about whether all of these things equal each other first so first of all let's just think about the value of the function there so remember we're doing Part A f of 0 is equal to we're going to use this first case because that's the case for X is less than or equal to 0 so f of 0 is going to be equal to 1 minus 2 sine I didn't write the two - sine of 0 - sine of 0 well sine of not theta sine of 0 is 0 2 times 0 0 so this whole thing is 0 1 minus 0 is 1 fair enough now let's think about the limit now let's think about the limit as X approaches 0 from the left hand side so as as we approach 0 from the left hand side of f of X so as we approach 0 from the left hand side we're dealing with values of X that are less than 0 so once again we're dealing with this first case up here so this is the limit as X approaches 0 from the left hand side of 1 minus 2 sine of X now sine of X is continuous it's a continuous function so this is going to be the same thing as 1 minus 2 sine of 0 which we've already figured out which is exactly equal to 1 so this is equal to 1 so the value of the limit as we approach from the left is the same as the value of the function now let's do it front as we approach from the right as we approach from values of X greater than 0 so let's think about the limit as we approach 0 from the right of f of X so here we're dealing with values of X that are larger than 0 so we're dealing with this case right over here so that's going to be the limit as X approaches 0 from the right of e to the negative 4x and for the exes that we care about or actually in general this is a continuous function this is a continuous function so it's this is going to be the same thing as e to the negative 4 times 0 which is just e to the 0 which is just 1 so this once again is equal to 1 so the function is equal to 1 at that point the limit as we approach from the left is equal to 1 and the limit as we approach from the right is equal to 1 so the function is continuous there
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