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Which of the followlng is an appropriate choice for & when proving the following limit? lim(-Tr - 6) =-4Assume > 0 is givenSelect the correct answer below;or...

Question

Which of the followlng is an appropriate choice for & when proving the following limit? lim(-Tr - 6) =-4Assume > 0 is givenSelect the correct answer below;or smalleror smallersmalleror smalleror smaller6 =or smaller

Which of the followlng is an appropriate choice for & when proving the following limit? lim(-Tr - 6) =-4 Assume > 0 is given Select the correct answer below; or smaller or smaller smaller or smaller or smaller 6 = or smaller



Answers

A sine limit It can be shown that $1-\frac{x^{2}}{6} \leq \frac{\sin x}{x} \leq 1,$ for $x$ near 0. a. Illustrate these inequalities with a graph. b. Use these inequalities to evaluate $\lim _{x \rightarrow 0} \frac{\sin x}{x}$.

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

Problem with 23 women as X approaches six of X squared minus six X over X squared minus five X minus six. Can't plug in six because we got X minus six in the Dominator. So we cancel that out with the top I factoring X in the top so we're left with X over X plus one thing. We plug in six to get 6/7.

In this problem were asked to determine which of the given inequalities has a solution set that spans from negative infinity to positive infinity. So since all of the quantities on the left hand side of our inequality R squared, that means that none of them should end up being negative. So this eliminates both our choices of B and D. We can also eliminate. We see because we noticed that it lacks an equal to component in the inequality. So through process of elimination, we see that option A has the desired solution, said.

We want to express the following limit is either negative infinity or positive infinity. So in this case are function given to us is four overhead minus six. Uh huh. And we're gonna be approaching six selfie approach six from the left. For example, We get a very small number but this number right here for me less than 605 5.9 5.99. So this number here to be negative. However, then we square it so it becomes positive. So because of the square value right here, any number in here, we'll end up getting an even smaller positive number, which means that this value right here for over the very, very small number tells us that this is going to go to positive infinity both directions. So the limit is positive infinity.


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