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Use the precise definition to compute the following limits_ Specifically, let € > 0 be an arbitrary small positive number: To show Jim_-- f(z) L; find posi...

Question

Use the precise definition to compute the following limits_ Specifically, let € > 0 be an arbitrary small positive number: To show Jim_-- f(z) L; find positive number & such that 0 < | - a| If(z) < 6 implies LI < € 6 will be expressed in terms oflim 8r + 4 = 44lim 31 + 2 = -1

Use the precise definition to compute the following limits_ Specifically, let € > 0 be an arbitrary small positive number: To show Jim_-- f(z) L; find positive number & such that 0 < | - a| If(z) < 6 implies LI < € 6 will be expressed in terms of lim 8r + 4 = 44 lim 31 + 2 = -1



Answers

Use Definitions $2.6 .1,2.6 .2,$ or 2.6 .3 to prove the given limit result. $$ \lim _{x \rightarrow-1}(x+6)=5 $$

Hello there. So here we have. The following limit is that involves an absolute value off X minus six. So let's evaluate this limit. This is almost trivial. This is just the absolute value of zero that is equals to zero. And this is or did I mean here. Then we need to use the EBS Lindell definition that say that for every epsilon greater than zero, the exist a delta greater than zero such that such that these difference that is going to be X 96 is greater than zero on a less them. This helps this delta. These implies that the difference between the function and the limit that in this case is just zebra is less than the absolute here. So what happened here is that we go to the limited zero, so we just have the absolute value of the function which, conveniently, in this case, is just the absolute value of the absolute value off. Okay, 96 on these should be less than done. Absolutely. But here's what we want. Actually, this is just the absolute value off X men. Six is less than Absalom. And here we have what we want. So We want something similar to this in order to find that delta that we need For every absolute on here, we have find it is just absolutely so. The Delta. He's just absolutely

Okay, so here we have the following function. That is just the constant function. So here the limit is trivial is just three. So this is early meat on. We can check, Actually, that, uh, under the Epsilon Delta definition, we require that for every Absalon greater than zero, there exists some the integrated zero such that, um this mhm this X in this case, minus six lessons from Delta implies that the difference between the function and limiting this cases three should be less. That's some, etc. For all etc. But here we have that or function is just three. So here we have three minus three, which is less than Epsom here. This is just syrup is less than absolute, but this is hold for every absolute for every delta, because for all X, the function is just three. So it doesn't matter that what delta we have here because always, we're going to obtain zero here in this part. And it is going to be zero, um, for all X. And this is less than Absalom. Just by the definition that we got here. So the answer here is that this is hold for every don't have greater down this year.

For the given function, I want to consider the limit as X approaches zero Of X -4. So that you do this, we're going to look at X -4 And we're going to approach zero in approaching zero, let's choose two different values that we can sort of bound instead of frame. Really given function that's going to be x equals 0.1. Mexico is -0.1. We do this. We see the corresponding why values Are negative 3.9 and -4.1. Sorry, delta Is equal to 0.1 And the excellent value, the Y value that is the distance from the desired limit is also .1. So based on this, we see that our adult the value equals are absolute value and that's regardless of how tight we make the window. That's because our graph has a slope of eat.

So for this problem we want to find the limit and uh filament as X approaches six of three, so we're gonna have the function F of X equals three. Now it's fairly obvious that this is a continuous function that always equals three. So the limit has X approaches any value is going to be three. Um But what we see is that regardless of what our delta value is, um we will be at that value of three, but just to show this we have f of six minus skip and then half of six. Let's see. And what we see is that regardless of what the C. Value is, it's going to equal three. Um So it doesn't matter what are delta value is any delta value will work. Um And that's if we let's see equal your 0.1, okay? Or even 100 No matter how big or small we spread out that built the value, we end up getting the same result.


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