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Current time:0:00Total duration:12:31

AP.CALC:

LIM‑1 (EU)

, LIM‑1.E (LO)

, LIM‑1.E.1 (EK)

multiple videos and exercises we cover the various techniques for finding limits but sometimes it's helpful to think about strategies for determining which technique to use and that's what we're going to cover in this video what you see here is a flowchart developed by the team at Khan Academy and I'm essentially going to work through that flowchart it looks a little bit complicated at first but hopefully it'll make sense as we talk it through so the goal is hey we want to find the limit of f of X as X approaches a so what this is telling us to do is well the first thing just try to substitute what happens when x equals a let's evaluate f of a and this flow chart says well if f of a is equal to a real number it's saying we're done but then there's this little caveat here probably and the reason why is that the limit is a different thing than the value of the function sometimes they happen to be the same in fact that's the definition of a continuous function which we talked about in previous videos but sometimes they aren't the same this will not necessarily be true if you're dealing with some function that has a point discontinuity like that or jump discontinuity or a function that looks like this this would not necessarily be the case but if at that point you're trying to find the limit towards the if as you approach this point right over here the function is continuous it's behaving somewhat normally then this is a good thing to keep in mind you could just see hey can I just evaluate the function at that at that a over there so in general if you're dealing with pretty plain vanila functions like an x squared or if you're dealing with rational expressions like this or trigonometric expressions and if you're able to just evaluate the function and it gives you a real number you are probably done if you're dealing with some type of a function that has all sorts of special cases and it's piecewise defined as we've seen in previous other videos I would be a little bit more skeptical or if you know visually around that point there's some type of jump or some type of discontinuity you've got to be a little bit more careful but in general this is a pretty good rule of thumb if you're dealing with plain vanilla functions that are continuous if you evaluate at x equals a and you get a real number that's probably going to be the limit but now let's think about the other scenarios what happens if you evaluate it and you get some number divided by zero well that case you are probably dealing with a vertical asymptote and what do we mean by a vertical asymptote well look at this example right over here we are just saying the limit then the darker color so if we're talking about the limit as X approaches 1 of 1 over X minus 1 if you just try to evaluate this expression at x equals 1 you would get 1 over 1 minus 1 which is equal to 1 over 0 which says ok I'm throwing it I'm falling into this vertical asymptote case and at that point if you wanted to just understand what was going on there or even verify that it's a vertical asymptote well then you could try out some numbers you could try to plot it you could say alright I probably have a vertical asymptote here at x equals 1 so that's my vertical asymptote and you could try out some values well let's see if X is greater than 1 the denominator is going to be positive and so my graph and you would get this from trying out a bunch of values might look something like this and then for values less than negative 1 or less than 1 I should say you're going to get negative values and so your graph might look like something like that until you have this vertical asymptote that's probably what you have now there are cases very special cases where you won't necessarily have the vertical asymptote one example of that would be something like 1 over X minus X this one here is actually undefined for any X you give it so it would be very you will not have a vertical asymptote but there's a very special case most times you do have a vertical asymptote there but let's say we don't fall into either of those situations what if when we evaluate the function we get 0 over zero and here is an example of that limit as X approaches negative 1 of this rational expression and let's try to evaluate it you get negative 1 squared which is 1 minus negative 1 which is plus 1 minus 2 so you get 0 in the numerator and in the denominator you have negative 1 squared which is 1 minus 2 times negative 1 so plus 2 minus 3 which is equal to 0 now this is known as indeterminate form and so on our flow chart we then continue to the right side of it and so here is a bunch of techniques for trying to tackle something in indeterminate form and likely in a few weeks you will learn another technique that involves a little bit more calculus called la Patel's rule that we don't tackle here because that involves calculus while all of these techniques can be done with things before calculus some algebraic techniques and some trigonometric techniques so the first thing that you might want to try to do especially if you're dealing with a rational expression like this and you're getting indeterminate form is try to factor it try to see if you can simplify this expression and this expression here you can factor it this is the same thing as X let's see X minus 2 times X plus 1 over let's see X what this would be X minus 3 times X plus 1 if what I just did seems completely foreign to you I encourage you to watch the videos on factoring factoring polynomials or factoring quadratics and so you can see here alright look if if I make the cat I can simplify this because as long as X does not equal negative 1 these two things are going to cancel out so I can say that this is going to be equal to X minus 2 over X minus 3 for X does not equal negative 1 I at sometimes people forget to do this part this is if you're really being mathematically precise this entire expression is the same as this one because this entire expression is still not defined at x equals negative 1 although you can substitute x equals negative 1 here and now get a value so if you substitute x equals negative 1 here even if it's formally formally taking it away to be mathematically equivalent this would be negative 1 minus 2 which would be which would be negative 3 over negative 1 minus 3 which would be negative 4 which is equal to 3/4 so if this condition wasn't here you could just evaluate it straight up and this is a pretty plain vanila function i wouldn't expect to see anything crazy happening here and if i can just evaluate it at x equals negative 1 I feel pretty good I feel pretty good so once again we're now going factoring we're able to factor we have value we simplify it we evaluate the expression the simplified expression now and now we were able to get a value we were able to get 3/4 and so we can feel pretty good that the limit here in this situation is 3/4 now let's and I would categorize what we've seen so far is the bulk of the limit exercises that you will likely encounter now the next 2 I was called slightly fancier techniques so if you get indeterminate form especially you'll sometimes see it with radical expressions like this rational radical expressions you might want to multiply by a conjugate so for example in this situation right here if you just tried to evaluate it x equals 4 you get the square root of 4 minus 2 over 4 minus 4 which is 0 over 0 so it's that indeterminate form and the technique here because we're seeing this radical and a rational expression and say let's maybe maybe we can somehow get rid of that radical or simplify it somehow so let's let me rewrite it square root of x minus 2 over X minus 4 let's take a conjugate let's multiply it by the square root of x plus 2 over the square root of x plus 2 once again I'm it's the same expression over the same expression so I'm not fundamentally changing its value and so this is going to be equal to well if I have a plus B times a minus B I'm going to get a difference of squares so it's going to be square root of x squared which is I'm just writing it's going to be square root of x squared minus 4 over well square root of x squared is is just going to be X minus four so let me rewrite it that way so it's X minus four over X minus four times square root of x plus two square root of x plus two well this was useful because now I can cancel out x equals 4 or X minus four right over here and once again if I wanted it mathematically to be the exact same expression I'd say well now this is going to be equal to 1 1 over the square root of x plus 2 for X does not equal 4 but we can definitely see what the function is approaching if we just now substitute x equals 4 into this simplified expression and so that's just going to be 1 over so if we just substitute if we just substitute x equals 4 here you get 1 over square root of 4 plus 2 which is equal to 1/4 and once again you can feel pretty good that this is going to be your limit we've gone back into the green zone if you were to actually plot this original function you would have a point discontinuity you would have a gap at x equals 4 but then when you do that simplification and factoring out that X my ORS or cancelling out that X minus 4 that gap would disappear and so that's essentially what you're doing you're trying to find the limit as we approach that gap which we got already there now this final one this is dealing with trig identities and in order to do these you have to be pretty pretty adept at your trig identities so if we're saying the limit as we do that in a darker color so for saying the limit as X approaches 0 of sine of X over sine of 2x well sine of 0 0 sine 0 0 is going to get 0 of 0 once again indeterminate form we've onto this category and now you might recognize this is going to be equal to the limit as X approaches 0 of sine of X we can rewrite sine of 2x as 2 sine X cosine X and then those two can cancel out for all X is not equaling for all X is not equal zero if you want to be really mathematical precise and so there would have been a gap there for sure on the original graph if you were to graph y equals this but now for the limit purposes you could say this is this limit is this limit is going to be the limit as X approaches 0 of 1 over 2 cosine of X and coasts and now we can go back to this green condition right over here because we can evaluate this at x equals 0 it's going to be 1 over 2 times cosine of 0 cosine of 0 is 1 so this is going to be equal to 1/2 now in general none of these techniques work and you will encounter a few other techniques further on once you learn more calculus then you fall on the baseline approximation an approximation you can do it numerically try values really really really close to the number you're trying to find the limit on you know if you're trying to find the limit as X approaches 0 try 0 point 0 0 0 0 0 0 0 0 0 1 try negative zero point zero zero zero zero zero zero one if you're trying to find the limit as X approaches four try four point zero zero zero zero zero one try three point nine nine nine nine nine nine nine nine nine and see what happens but that's kind of the last ditch the last-ditch effort