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Evaluate each LimitpointVx _ 2 lim x -4 x-41/41/2...

Question

Evaluate each LimitpointVx _ 2 lim x -4 x-41/41/2

Evaluate each Limit point Vx _ 2 lim x -4 x-4 1/4 1/2



Answers

Evaluate the following limits $\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}$

In this exercise. We want to evaluate the limit so we can do is first see if we can use will be tells rule. So we would do that by seeing if we get an instrument form when we have ex approach for for what we have here. So if you have X approach for in one over the square root of X minus two, we're going to get infinity because we're just going to have a denominator that is gonna be really, really close to zero. But the new wave in the same So the numbers just gonna keep getting bigger. Mayor's approaching infinity. Then we have this minus sign right here. Then we have four over X minus four. So if you have experts for here, we get the same thing. We have the numerator say the same of the denominator, get closer and closer to zero. So the number is gonna get really large still have infinity. My sin city, which is in fact indeterminant so we can use will be tells. So we're going to do is we're going to first combine this into a single fraction. And one thing that's really important to note is that if we have the square root of X plus two times the square root of X minus two, they get X minus four. So what we can do to get X minus four And this denominator is multiplied by the square root of exposed to over the skirt of extras too. So we get the square of explosives to over X minus four minus four over. Excellent is for which is what we're trying to limit up. And this ends up being the square root of X minus two over X minus floor soul. Right. This again Here we'll have the limit of ex approaching for this is just by combining our fractures right now square root of X minus two over X minus four. So we know, Of course we can use will be tile on this. So we're gonna say that this is a of X. This is G of X. So if we take the derivatives, we have the derivative of X and the derivative G of X. Of course you have extra isn't very easy is just going to be one, and the derivative of X is just going to be bringing power down of the square root of X number of X to the negative on half. So now we're trying to take the limit of ex approaching four of the derivative F over the road of G. So just one. So we divide me one. So we'll have one over two square root of X and we can see that if X is approaching four. Here we get one over two times, but the four into the square root, we get two times two. So our answer is put a different color. One over four. Yes, I do it.

So here we're looking at the limit as X goes to two, X squared minus four divided by x minus two. Um If we try to directly substitute into into this function since our numerator in our denominator, both polynomial functions, we would be allowed to do that. However if we do we're going to see that this is gonna be four minus four divided by two minus two, Which is equal to zero, divided by zero. So we can say that this limit might exist because we have zero divided by zero and not um some other number divided by zero. And so what we want to do now is we want to try and factor dysfunction in a way that will let us um actually cancel out this X -2 term down here. And so Since the numerator is a difference of squares, we actually can factor the numerator into X-plus two Times X -2. And then we're going to see that this X -2 term in the numerator and the denominator are going to cancel. And so this is going to be equal to the limit excuse to to just X plus two, which now we can directly substitute into for X and so to speak, with a two plus two, Which is equal to four.

So what we wanna do to figure out this limit is we're gonna want to figure out how can we actually go about um canceling this x minus four um denominator term down here and the way that we can actually do that is by multiplying by the conjugate of our numerator, so by multiplying by the square root of X plus two, divided by the square root of X plus two. And so here what's going to happen is the numerator is going to turn into, go ahead and write this stuff down. It's going to be squared of X times squared of X which is just X. And then the squared of x minus two which is negative two times squared of X. And then we have um squared of X plus are multiplied by um positive too. So be plus two times a squared of X. And then we have minus four and our numerator or denominator is still x minus four times the square root of X plus two. And so we can see that these two terms negative two times a squared of x and positive two times a squared of X are going to cancel. And our numerator is now going to become just X -4. And so we'll be able to cancel that out with this X -4 terms in the denominator. And so then we're just going to be left with the limit As X goes to four. So this he calls the limit X goes to four of one divided by the square of X plus two. And now we can directly substitute in four. And so this is gonna be one divided by the square to four plus two which is equal to one divided by two plus 2 Which is equal to 1/4. So here again, what we had to do was multiply um the numerator and the denominator by the conjugate of the numerator, which then allowed us to cancel out the term in the denominator that was um causing this limit to be undefined and we were able to find the actual value of the limit.

Okay, so we're looking at this limit as X goes to two of x divided by x squared minus four. Um What we can do first is we can factor this numerator or this denominator and so this will be equal to the limit as X goes to two of X divided by X plus four times X. We're sorry not expose for exposed to Times X -2. And we can see that we're not going to be able to actually um cancel out this x minus two factor. We could minus two and then plus two and then split up this fraction but then we're still going to run into the same problem so that's not really going to work here. And so what we're going to figure out is that when we take this limit we're going to get the limit of the numerator would be equal to two divided by zero. So this is going to be equal to two divided by zero. Which means that this limit does not exist since we have a number that is not equal to zero divided by zero when we take this limit. And that means that this limit here isn't going to exist. So we can say that this does not exist.


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