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Current time:0:00Total duration:9:32

AP.CALC:

LIM‑1 (EU)

, LIM‑1.E (LO)

, LIM‑1.E.1 (EK)

let's see if we can find the limit as X approaches negative 1 of X plus 1 over the square root of x plus 5 minus 2 so our first reaction might just say okay well let's just use our limit properties a little bit this is going to be the same thing as the limit as X approaches negative 1 of X plus 1 over over the limit the limit as X approaches negative 1 of square root of x plus 5 minus 2 and then we could say all right this thing up here X plus 1 if this is if we think about the graph y equals x plus 1 it's continuous everywhere especially at x equals negative 1 and so to evaluate this limit we just have to evaluate this expression at x equals negative 1 so this numerator is just going to evaluate to negative 1 plus 1 and then our denominator square root of x plus 5 minus 2 isn't continuous everywhere but it is continuous at x equals negative 1 and so we can do the same thing we can just substitute negative 1 for X so this is going to be the square root of negative 1 plus 5 minus 2 now what does this evaluate to well in the numerator we get a 0 and in the denominator negative 1 plus 5 is 4 take the principal root is 2 minus 2 we get 0 again so we get we got 0 over 0 now when you see that you might be tempted to give up you say oh look I'm there's a 0 in the denominator maybe this limit doesn't exist I'm maybe I'm done here what do I do and if this was nonzero up here in the numerator if you're taking a nonzero value and dividing it by a 0 that is undefined and your limit would not exist but when you have 0 over 0 this is indeterminate form and it doesn't mean necessarily that your limit does not exist and as we'll see in this video in many futures 1 the future ones there are tools at our disposal to address this and we will look at one of them now the tool that we're going to look at is is there another way of rewriting this expression so that we can evaluate its limit without getting the 0 over 0 well let's just rewrite let's just take this let me give it so let's take this thing right over here now let's say this is G of X so essentially what we're trying to do is find the limit of G of X as X approaches negative 1 so we could write G of X is equal to X plus 1 and the only reason why I'm defining it is G of X is just to be able to think of it more clearly as a function and manipulate the function and then think about similar functions over X plus 5 minus 2 or X plus 1 over the square root of x plus 5 minus 2 now the technique we're going to use is when you get this indeterminate form and if you have a square root in either the numerator denominator it might help to get rid of that square root and this is often called rationalizing expression in this case you have a square root in the denominator so it would be rationalizing the denominator and so this would be the way we would do it is we'd be leveraging our knowledge of difference of squares we know we know that a plus B times a minus B is equal to a squared minus B squared you learn that in algebra a little while ago or if we had the square root of a plus B and we were to multiply that times the square root of a minus B well that'd be the square root of a squared which is just going to be a minus B squared so we can just leverage this these ideas to get rid of this radical down here the way we're going to do it is we're going to multiply the numerator and the denominator by the square root of x plus 5 plus 2 all right we have the minus 2 so we're multiply it times the plus 2 so let's do that so we have square root of x plus 5 plus 2 now we're going to multiply the numerator times the same thing because we don't want to change the value of the expression this is 1 so if we take the expression divided by the same expression it's going to be 1 so this is so square root of x plus 5 plus 2 and so this is going to be equal to this is going to be equal to X plus 1 times the square root times the square root of x plus 5 plus 2 and then the denominator is going to be well it's going to be X the square root of x plus five squared which would be just X plus five and then minus two squared minus four and so this down here simplifies to X plus five minus four is just X plus one so this is just this is just X plus one and it probably jumps out at you that both the numerator and the denominator have an X plus one in it so maybe we can simplify so we could simplify and just say well G of X is equal to the square root of x plus five plus two now some of you might be feeling a little off here and you would be correct your spider senses would be say oh is this is this definitely the same thing as what we originally had before we cancelled out the X plus ones and the answer is the way I just wrote it it is not the exact same thing it is the exact same thing everywhere except at x equals negative one this thing right over here is defined at x equals negative one this thing right over here is not defined at x equals negative one and G of X was not was not so G of X right over here you don't get a good result when you try x equals negative one and so in order for this to truly be the same thing as G of X the same function we have to say for X not equal to negative one now this is a simplified version of G of X it is the same thing for any of input X that G of X is defined this is going to give you the same output and this has the exact same domain now now that we put this constraint in as G of X now you might say okay well how does this help us because we want to find the limit as X approaches negative 1 and even here I had to put this little constraint here that X cannot be equal to negative 1 how do we think about this limit well lucky for us we know lucky for us we know that if we just take another function f of X if we say f of X is equal to the square root of x plus 5 plus 2 well then we know that f of X is equal to G of X for all X not equal to negative 1 because f of X does not have that constraint and we know if this is true of - if this is true of two functions then the limit as X approaches the limit let me write this down since we know this because of this we know that the limit of f of X as X approaches negative 1 is going to be equal to the limit of G of X as X approaches negative 1 and this of course is what we want to figure out was the beginning of the problem and but we can now use f of X here because only at x equals negative 1 that they are not the same and if you were to graph G of X it just has a it has a point discontinuity or removable disk on tour or I should just say oh yeah a point discontinuity right over here at x equals negative 1 and so what is the limit and we are in the homestretch now what is the limit of f of X or we could say the limit of the square root of x + 5 + 2 as X approaches negative 1 well this expression is continuous it or this function is continuous at x equals negative 1 so we can just evaluate it at x equals negative 1 so this is going to be the square root of negative 1 plus 5 plus 2 so this is 4 squared principal root of 4 is 2 2 plus 2 is equal to 4 so since the limit of f of X as X approaches negative 1 is 4 the limit of G of X as X approaches negative 1 is also 4 and if this little this little I guess you could say leap that I just made right over here it doesn't make sense to you think about it think about it visually think about it visually so if this is my y-axis and this is my x-axis G of X looked something like this G of X G of X let me draw it G of X looked something something like this and it had a gap at negative one so it had a gap right over there while f of X f of X would have the same graph except it wouldn't have it wouldn't have the gap and so if you're trying to find the limit it seems completely reasonable well let's just use f of X and evaluate what f of X would be to kind of fill that gap at x equals negative 1 so hopefully this graphical version helps a little bit or or if it confuses you ignore it

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