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In using the € ~ 6 definition of limit to prove that Hn (6x ~ 1) 4, what is the largest thet 6 can be for &n arbitrary € > 0? A;B, beC, 56 - 1D, ...

Question

In using the € ~ 6 definition of limit to prove that Hn (6x ~ 1) 4, what is the largest thet 6 can be for &n arbitrary € > 0? A;B, beC, 56 - 1D, 6

In using the € ~ 6 definition of limit to prove that Hn (6x ~ 1) 4, what is the largest thet 6 can be for &n arbitrary € > 0? A; B, be C, 56 - 1 D, 6



Answers

(a) What are the possible values of / for $n=6 ?$ (b) What are the possible values of $m_{l}$ for $l=6 ?(c)$ What is the smallest possible value of $n$ for which $l$ can be $4 ?$ (d) What is the smallest possible $l$ that can have a $z$ component of $4 \hbar ?$

Okay, so we're wanting to know what is a of six well off of six. We have to first to sell which function will be closing it into. And since six is, um, you know, greater than or equal to six we have to play into the bottom one. So then we have f of X busy with negative 1/3 X minus two because that top function of three X plus one we only use it when the excess lesson six which is not the case here. So then if I put that in, FF six will be united of 1/3 time six minus two. So f of six is going to equal negative two minus two were just negative for.

What are the possible values of Elwin and is equal to 6? So L max is equal to you And uh -1 which is equal to five. So that means l. 01234 five. For any equal to six Part B. What are the values of impossible? When L is equal to 6? That is plus or -6. Plus or -5? Plus or -4? Plus or -3. Plus or -2. Closer -1 and zero. What is the smallest possible value of N. For which el can be for? If L can before we have that N is greater than or equal to our plus one Which is equal to five. So the smallest possible end is five. What is the smallest possible L. That can have a Z. Component of four? H. Bar. So if it can have a value of four h. Bar. EMC Bell is equal to four, which means L is greater than or equal to four. So the smallest possible value of L. Is boy.

May. Inequality is a greater than or equal to negative six. I will do a number line to shade negative six is on the left and zeros on the right says, greater than or equal to, which is a closed circle. Greater means to the right and the aero shading that I will check to see replacing the value of a negative six. Is it greater than or equal to negative? Six. And it's not greater, but it's equal. So it is a solution, on top of the fact that it's a closed circle, then negative 6.1. Is it greater than or equal to negative? Six? And the answer is no, because negative 6.1 is on the left hand side of the negative six. Next value wear testing. Negative 5.9. Is it greater than or equal to negative six. And the answer is yes, because that is on the right hand side of negative

So here we have the limit of the absolute value of two x as X. A purchase negative three equals six. We went to prove it. We will use the definition and we'll start with that. So we have. Zero is less than the absolute value of X minus negative. Three. His lesson. Delta. And you'll see why we leave that as negative three in a bit. And then we have the absolute value. The absolute value of two X minus six. He's less than absolute. We'll start with Delta. We're gonna use this identity here to help rewrite Delta. So we have. Zero is less than absolute value of the absolute value of X minus the absolute value of negative three, which is less than what are original inequality Waas, which is X minus negative three. Or rather, it's less than or equal to, and that's less than Delta. Because of this inequality, we can simplify this little and just say that zero is less than the absolute value we're going to use this. The absolute value of negative three is three, so we have the absolute value of the absolute value of X minus three. There's less than Delta, and this is the inequality we're gonna use late. So let's go over to our Absalon. We can factor this two out of this absolute absolute value, getting us two times the absolute value of X minus six. And then we can factor to out of this absolute value and get two times the absolute value. The absolute value of six X minus three, his less than absolute. If we divide both sides by two, we get the absolute value of the absolute value of X minus three is less than epsilon over to. We now have matching relationships for Delta and Epsilon over to so we can say that Delta equals Absalon over to


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