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Let f(r) = 4r - 1 We will show linf(z) = 7 with the 2-6 defnition 0f & limit when 2 and L = 7. That aim to prOve that for every € > there is & 6 &g...

Question

Let f(r) = 4r - 1 We will show linf(z) = 7 with the 2-6 defnition 0f & limit when 2 and L = 7. That aim to prOve that for every € > there is & 6 > 0 such that whenever <k-el 6 it folloxs that Vfl-)- L = 0.1, find that whenever <hr-2 < 6 we get [fls) -7< For _0.01 , findthat wbenever <hr -2 < 6 we get Ifls) -7<For € > find $ (in terms of €) so that whecever 0 <k-2 < & we get Ul-)-I< Using this procesS ,say that for Ctery € > there is

Let f(r) = 4r - 1 We will show linf(z) = 7 with the 2-6 defnition 0f & limit when 2 and L = 7. That aim to prOve that for every € > there is & 6 > 0 such that whenever <k-el 6 it folloxs that Vfl-)- L = 0.1, find that whenever <hr-2 < 6 we get [fls) -7< For _ 0.01 , find that wbenever <hr -2 < 6 we get Ifls) -7< For € > find $ (in terms of €) so that whecever 0 <k-2 < & we get Ul-)-I< Using this procesS , say that for Ctery € > there is & 6 > namely such that whenever 7<6 = If(z) Thus wee have used the €-& definition of limits shox (Using the €-& definition limits inctions that are not linehr depends in more complicated way on and f(r) ) harder btrause



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Limit proofs Use the precise definition of a limit to prove the following limits. Specify a relationship between $\varepsilon$ and $\delta$ that guarantees the limit exists. $\lim _{x \rightarrow-3}|2 x|=6(\text { Hint: Use the inequality }\|a|-| b\| \leq|a-b|$ which holds for all constants $a$ and $b$ (see Exercise 74 ).)

For the given problem. We want to use the formal definitions of limits to better understand and improve the given limit. So we're focusing on the Function three x plus seven. We want to show what happened When X approaches zero from the right and left. Well to show this, we consider what the graph is going to look like. If we have x equals 7.1 Max equals 6.9. So when we do this actually this x equals 0.1. Okay -0.1. We do this. We see that the corresponding epsilon value is going to be three times The original delta value. So we have our adult a value of 0.1 and the epsilon value, which is the value above the limits of 7.3 minus the limit of seven, Appreciate it absolutely is going to be your .3. That means that we have our realtor delta value equal epsilon over three.

For the given problem we want to prove um The limit using are more technical formal definitions of the limit. So here we have the function X-plus six. Using our formal definitions, we want to show how It's going to equal five when X approaches -1. So let's approach -1 in the graph. Consider What happens if we have x equals a negative 1.1 and X equals a negative 0.9. Here we see that there is a .1 difference or adult a value 0.1. That's the difference in the X value from this point that we're looking for. But we and we see that the difference in um from the actual limit which is five, the difference from that point is going to be 0.1 as well. That's your excellent value is 0.1 with that and I'll agree to that depth delta equals epsilon. And that will ensure that our interval stays um such that we've proven the limits.

Alright for Low P tells rule, We gotta make sure that we have the indeterminate forum in this case of 0/0. But there are other Indeterminant forms. Uh, s Oh, that's my first step is to show you the limit as r approaches, one of a times are to the end minus one all over r minus one. So what you need to do is to show the work that when you plug in one and for our well, one toe any power still one minus 10 We're looking at a time zero, which equals zero. So we get zero on top. We're looking the denominator 01 minus one is zero. We get zero on bottle. So that's telling us, Hey, we have differential functions. We have 0/0 so we can move forward. And what Loki tells rule tells us is to take the derivative of the top. Now it's technically the product rule, but it is just a constant. So if I were to distribute that in, it's just a constant eso we're looking at a and times are to the n minus one power because remember the derivative we bring the end in front. Subtract one from the experiment. Uh, on then minus a again, I'm distributing the derivative of a zero because it's a constant over the derivative of the bottom. The director of our is one eso. Now I can plug one and for our So we're looking at a times end times one to the n minus one Power. Now I remember one toe. Any power, no matter what is is one. So eight times and times one is just eight times n or that's your final answer and yeah.

To the following problem. We want to use the definitions to prove given limit result. So the function we have here is the limit as X purchased through our function is F of X equals mhm. Spread -7, experts. 12. Yeah. Okay. And then this is going to be divided by X -6. Now it's important that we keep in mind the fact that um we can factor this out and we're gonna spend more time on this problem factoring natural limit because there's a lot of examples of how to take the limit. However, there's not a lot of examples of how to factor. So it's an important skill to remember. Keep in mind you can factor out of two here making this X -3 And the reason you do that is because this right here is equal to X -3. Time to X -4. Then you can factor out this X -3. Now what we have is x minus 4/2, Which is going to be 1/2 acts thanks to making it much more easy to determine your excellent value.


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