5 answers

3) The Limit isNow explain why:

Question:

3) The Limit is Now explain why:


Answers

Explain why the limits do not exist. $$\lim _{x \rightarrow 0} \frac{x}{|x|}$$

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So why did the limits at Infinity work? Well, one thing to note is that the symbol for Infinity is just referring to really large numbers. So if all the operations with limits work with small numbers certainly listen worked with large numbers, like a 1,000,010 million, 100 million, Since they're all just numbers, all the operations that we've been working with throughout all our lives, it will still apply to these big numbers, so nothing really changes..

Okay, so today we're going to solve the problem. The limit as X goes to three of two x plus X, the absolute value of X minus three. So in order for me to solve this limit outbreak Li I first need to determine does the limit exist at two X and at the absolute value of X minus three. So I'm going to use my graphing calculator to help me determine that. So I've already graphed both of these problems for both of these functions.

Um, so our first function f of X equals two exes graft in red, and we can see that X equals three would be right here. Um, so from the left and the right, in order for this limit to exist, um, it needs to be approaching the same number, um, as X approaches three from the left and the right. So as X approaches three from the left, you can see it's getting bigger and bigger and bigger until it approaches. Six. As X approaches three from the right, it's getting smaller and smaller and smaller until our function approaches six as well here.

So our limit does exist at f of X equals two X um, which makes sense, since this is a just a simple linear equation. So now let's talk about, um, the absolute value of X minus three, which is this blue graph. So down here is where X is equal to three. We can see from the right, um, as X approaches three, why is approaching getting smaller and smaller and smaller approaching zero and from the left as X approaches three. Why is getting smaller and smaller and smaller and smaller until it approaches zero? So it does approach the same number from the left and the right, meaning that our limit exists for both of these functions, so we can find our limit as X approaches.

All right, so I'm going to do that algebraic lee, and I'm going to use the rule that tells me my limit law. That tells me the limit as X approaches A of X is equal to basically that just says that I can plug in this number right here. The number that I'm trying excite. I'm getting X to approach into both of these functions. Um, and whatever that is equal to that is our limit.

At that point, at X equals three. So that tells me that the limit as X approaches three of two x plus, the absolute value of X minus three is equal to two times three plus the absolute value of three minus three. Sorry for all of these little scribbles. Okay, so now two times three is equal to six, plus the absolute value of three minus three, which is zero. The absolute value of zero is zero.

So six plus zero is six. That means my final answer is yes. This limit exists and it is equal to six..

Okay for this problem when taking a look at rule four, which simply says, if I have the limits of a fraction right here. So I have some function in the numerator and a different function in the denominator. Then I can just go ahead and take the limit of the numerator divided by the limit of the denominator. Um, so if those limits exist and we have a over B than the limit of the fraction is just the fraction of the limits. So essentially we can treat the numerator and denominator separate.

Now, we have to be very careful because it only works as long as the denominator is not going to zero. And in the hypothesis, it said that these limits do have to exist, So we can't use this or we shouldn't use, um if these limits are infinite, so essentially limits distribute over fractions, but only if the denominator isn't going to go to zero.

To find a limit as X approaches. Three the limits. X approaches three of two X close. Absolute value X minus three. Right.

So to do this, you just have Teoh. I just have to substitute whoever easy exit. Just with three. Does it? That is gonna be two times three plus absolute value of three minus three. This is zero.

Right? So this is just six. So six is the answer to the limit exist and the limit is equal to six..

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