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Mathematics 14 Online
OpenStudy (stephy321):

help!! stuck between C and D .

OpenStudy (stephy321):

these are the answer choices!

OpenStudy (zzr0ck3r):

what is the function?

OpenStudy (mathmate):

Two questions: 1. What is the given f(x), must have been defined higher up in the question. 2. If you are hesitating between c and d, can you explain the hesitation(s)? what supports that there are two values of k?.... and the same question for three values.

OpenStudy (stephy321):

The f(x) is attached below. I'm confused between C and D because when I go through y=1, I see that there is an open dot at (1,1) and at (4,1) and (6,1) there are actual values. I'm confused with (1,1) because I do not quite remember if the limit cares whether their is a closed or open circle unlike f(x). (I tried explaining the best I could...)

OpenStudy (stephy321):

@mathmate

OpenStudy (mathmate):

You may have posted the wrong image. It is exactly the same as what you posted earlier. We still do not know if f(x) is a function, whether it has a limit, or how it is defined. You are describing an image that we don't see...lol

OpenStudy (stephy321):

That's the comptete problem

OpenStudy (stephy321):

complete**

OpenStudy (stephy321):

this is the graph?

OpenStudy (mathmate):

Thank you for posting the image of the function. Without it, we'd be making a guess, which should NOT happen when we're working with math. The point (1,1) is an open dot, which means that f(1) does not exist...nor is the point (1,1) part of the curve. To answer this question, we must know the conditions of the existence of a limit. For a limit to exist at x->k, BOTH of the following conditions must be satisfied: 1. f(k) must exist and be finite. 2. Limits from the left (x=k-) and from the right (x=k+) must both exist and are both equal to f(k). Since f(1) is an open circle at (1,1), and f(1) is not defined elsewhere with a closed circle, we conclude that f(1) does not exist, and therefore the limit does not exist [using condition 1 above], despite partial satisfaction of condition #2 above. For future reference, read the following link: http://www.mathwarehouse.com/calculus/limits/how-to-determine-when-limits-do-not-exist.php

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