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# How would you "remove the discontinuity" of $f$? In other words, how would you define $f(2)$ in order to make $f$ continuous at 2?$f(x) = \dfrac {x^3 - 8}{x^2 - 4}$

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We have f of x equals execute -8 over X squared minus four. Dysfunction is discontinuous when X equals two, So x cannot equal negative two or 2 for this function if X was negative two or two, then the denominator export minus four would come out to be zero and you can't divide by zero. So our function F of X is discontinuous at these values of X. Specifically, we're going to look at when X is too, so this function is discontinuous when X is too. No, we want to see how we can define F of two. Uh so that we can, you know, remove the discontinuity, we know that this function is discontinuous at two. Can we redefine uh F of two? In other words, can we say f of two equals some number So that this function becomes continuous at two. Well, function is continuous as long as the limit. Um let me rephrase that uh this function when we redefine it, of course with FF two a function is continuous. Uh Two. If the limit of the function as X approaches to is equal to to function at two. So in order to find out what we want to define F of two, F at two equal, we want to find out what is the limit of our function F of X as X approaches to. Okay, so we want to find the limit of F of X as X approaches two. Now it's not as uh simple as trying to directly substitute to in forex. If you want to find a limit of this function. Okay, the limit of F of X as X approaches to, it's not as simple as just as as just directly substituting to win for X because if you substitute to infer x, um then to square minus four is zero and you can't divide by zero. Uh And of course if you substitute to infer x, the numerator is also zero. Okay, so if we try directly substituting to infer X two cubed minus 802 squared minus four is zero. And you get the indeterminate form of a limit when you get 0/0, so directly substituting to infer X is not going to work. So how will we find the limit of this function F of X. Okay, how do we find the limit of F of X as X approaches to without directly substituting to win for X? We will do it by factoring So executed -8 factors um Into X -2 -2 times X squared plus two X plus four. So execute -8 factors into X -2 times X squared plus two, X plus four, X squared minus four factors into x minus two Times X-plus two. Now, since X is approaching to but does not become too we don't have to worry about x minus two being to minus two, which is zero. So you don't have to worry about this factor being a zero in the denominator because X approaches to it does not ever he come to it doesn't get To to, it just approaches to. So what we're going to do is since we're multiplying by X -2 in the numerator and the denominator, we can cancel those factors. No um F of x t executed minus eight over x squared minus four. We rewrote it in factored form, so execute minus eight over x squared minus four. Was this when we factored the numerator and the denominator um of F of X. And when we factored it, we found that there were common factors that we were able to catch. So the limit of this function as X approaches to is really now the limit of what's remaining as X approaches to. Uh well, this is a continuous function at two. This is a continuous function at to another way I can say that is uh the limit of this polynomial up top exists as X approaches to and the limit of this polynomial in the denominator exists as X approaches to. And so we can simply now just plug to infer x everywhere. You see X to calculate this limit. So the limit of this expression as X approaches to is two squared Plus two times 2 plus four over X plus two. As extra approaches to X plus two approaches two plus two. So up top two squared is four plus two times to another four plus another four. We have 12 in the numerator. Okay, foursquare plus four plus four. two squared is four plus four plus four is 12 divided by four. So 12 divided by four is straight. So the limit of our function as X approaches to is three. So if we want to function F of X, if we want to redefine uh f of X, um or specifically if we want to define F of to to be a certain number, so that F is continuous at two. Well, here's what it means for the function F of X to be continuous. Set to F of X will be continuous at x equals two. As long as the limit of F of X as X approaches to is equal to the value of F at two. F is continuous at two, if the limit of F of X as X approaches to equals the value of F at two. While the limit of F of X as X approaches to is three. So if we make f of two equal to three than we are guaranteeing continuity at two because the limit of F F X as X approaches to will equal the value of F of to the limit of F of X as X approaches to is three. Where you are defining the value of F at 22. B three. So now the limit of our function as X approaches to is equal to the value of our function at two. So now, as long as we define ff two to equal three, we just removed the dis continuity that we previously had it, too, and we now have a newly defined function F of X that is continuous when X is too.

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