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anonymous

  • one year ago

Are these correct?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Nnesha @ganeshie8

  3. anonymous
    • one year ago
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    4 is correct.

  4. anonymous
    • one year ago
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    How is that one correct? I got the same answers as the first 3 problems but not number 4.

  5. anonymous
    • one year ago
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    I believe your goal is to plug in x, the x is stated for you. (2)^2 + 5(2) + 6 -------------- (2)+2 4+10+6 --------- 4 20/4 = 5. Doing this in my head, apologies for any error.

  6. anonymous
    • one year ago
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    Oh I made a mistake in that. Thanks! Can you explain number 5 too @Mikeyy1992?

  7. anonymous
    • one year ago
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    I shall try, one sec.

  8. anonymous
    • one year ago
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    Okay

  9. anonymous
    • one year ago
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    x-1 ------ x^2 - 1 X is stated to be 1. 1-1 ------ 1^2 - 1 0 - 1*1 - 1 0 - 0 The answer shall be 0, if I am not mistaken.

  10. anonymous
    • one year ago
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    Okay I got that too but my teacher put 1/2.

  11. anonymous
    • one year ago
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    Do you know how to solve 6 and 7? Sorry if I'm asking for too much.

  12. anonymous
    • one year ago
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    @Mikeyy1992

  13. anonymous
    • one year ago
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    I can do them in my head, but I do not know how to fully explain 6 and 7, but from my knowledge, the answers are both correct. For 5, I have dealt with the same exact puzzle, so I know the answer is 0, anyways.

  14. anonymous
    • one year ago
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    Can you write the steps on how you solve it? I'll see if I can understand when I look at the steps.

  15. anonymous
    • one year ago
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    6, already proves it will be DNE, with the second statement of x>2. With number 7, I see cos, so already think negative wise, from past learnings, which I know by heart. It's more so, making e to -e then x/3 into a factor which produces -e^3. I am not good at explaining some terms in mathematics, apologies.

  16. anonymous
    • one year ago
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    Why does the second statement x>2 prove that it's DNE?

  17. anonymous
    • one year ago
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    Zepdrix, might point you in a better direction than me; if I am making any mistakes.

  18. zepdrix
    • one year ago
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    Oooo it's Princess Jasmin!! :D

  19. zepdrix
    • one year ago
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    Are these answers that the teacher posted? Or are we checking YOUR answers?

  20. anonymous
    • one year ago
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    No it's my teacher's answers and haha ya that's me! :)

  21. zepdrix
    • one year ago
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    For number 6, remember that for a limit to exist `at a particular point`, The limits from each direction must match \(\large\rm x\to2^{+}\) needs to give the same as \(\large\rm x\to2^{-}\)

  22. anonymous
    • one year ago
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    Gosh, thank you for stating that way better than me, zepdrix! Knew you could! (:

  23. zepdrix
    • one year ago
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    \[\large\rm \lim_{x\to2^{-}}2x^2-4x\]So if you look at this big ole piece-wise function... when we approach 2 from the `left side`, we're using the top piece, understand why? :o

  24. anonymous
    • one year ago
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    The top piece meaning 2x^2 -4x?

  25. zepdrix
    • one year ago
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    ya

  26. anonymous
    • one year ago
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    We are trying to see if the function is indeterminate or if the point doesn't exist? Or if the function is continuous or if it has a hole?

  27. zepdrix
    • one year ago
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    For piece-wise functions, we need to make sure that the `limit from the left` agrees with the `limit from the right`. If they don't agree, then the limit does not exist at that point. So we have some big piece-wise function, let's call it f(x). We want to know whether or not \(\large\rm \lim_{x\to2}f(x)\) exists, and if so, find the value. To determine this, we need to check and see if the left and right limits agree,\[\large\rm \lim_{x\to2^-}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]

  28. zepdrix
    • one year ago
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    So again I'll ask,\[\large\rm \lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}2x^2-4x\]Do you understand why I'm replacing f(x) with this top line in the piece-wise function for x approaching from the left?

  29. anonymous
    • one year ago
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    Okay so wouldn't you just insert in a 2 on both sides of the equation? It would still give the same answer right?

  30. zepdrix
    • one year ago
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    Yes, if that's easier, maybe just think of it that way :) The pieces need to give the same output at x=2.

  31. zepdrix
    • one year ago
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    \[\large\rm 2x^2-4x \stackrel{?}{=}4\sin\left(\frac{\pi x}{4}\right)\]So are these equal when you plug in x=2? :)

  32. anonymous
    • one year ago
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    0 = 3.99999

  33. anonymous
    • one year ago
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    So no it doesn't match.

  34. anonymous
    • one year ago
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    Oh so do you that process for all inequalities that pop up with the lim function?

  35. zepdrix
    • one year ago
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    These were our steps: \[\large\rm \lim_{x\to2^-}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)\]\[\large\rm \lim_{x\to2^-}2x^2-4x\stackrel{?}{=}\lim_{x\to2^+}4\sin\left(\frac{\pi x}{4}\right)\]\[\large\rm 2(2)^2-4(2)\stackrel{?}{=}4\sin\left(\frac{4\pi}{4}\right)\]\[\large\rm 0\ne4\]I suppose yes. It might get a tiny bit more complicated if your teacher throws piece-wise functions at you with 3 or 4 pieces. But it should always be the same process :O

  36. zepdrix
    • one year ago
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    Should we go over number 5? :o looks like you were both a little confused on that one.

  37. anonymous
    • one year ago
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    Ya can you go over that and number 7? Sorry!

  38. zepdrix
    • one year ago
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    \[\large\rm \lim_{x \to1}\frac{x-1}{x^2-1}=\frac{0}{0}\ne0\]0/0 is an indeterminate form, it's not equal to 0. So we have to work our way around it some how.

  39. zepdrix
    • one year ago
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    Notice that the denominator is the difference of squares,\[\large\rm x^2-1=x^2-1^2\]We have a rule for factoring the difference of squares, do you remember it? :)

  40. anonymous
    • one year ago
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    So we have to multiply both top and bottom by x+1?

  41. zepdrix
    • one year ago
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    Umm sure, that's another option! That's pretty much the same thing I was doing, but in reverse.

  42. anonymous
    • one year ago
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    Oh okay. That's how my teacher taught me. So after reducing or crossing out you would get 1/(x+1)?

  43. anonymous
    • one year ago
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    And then you would have to plug in 1 right?

  44. zepdrix
    • one year ago
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    Yes :)

  45. anonymous
    • one year ago
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    Oh, that's what I forgot, been a couple years! :P Thank you again, Zepdrix.

  46. anonymous
    • one year ago
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    Ohhhh that's why the answer is 1/2!

  47. zepdrix
    • one year ago
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    So this is what you do with limits. Step 1: You try to plug the number directly in. ~If you run into trouble, you have to try and `cancel stuff out`. Step 2: Cancel stuff out. Step 3: Try to plug the number directly in again.

  48. anonymous
    • one year ago
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    Okay I understood that. Thanks! Can you help me with number 7?

  49. zepdrix
    • one year ago
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    See the steps I listed? Number 7 is going to be a piece of cake for us. We'll apply step 1, and we won't run into any problems. It will simplify down to an answer.

  50. zepdrix
    • one year ago
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    Plug 3 directly into the function. We get something like this,\[\large\rm e^{3}\cos\left(\frac{\pi(3)}{3}\right)\]yes?

  51. zepdrix
    • one year ago
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    And now simplify :)

  52. anonymous
    • one year ago
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    \[e^3\cos(3.14) = e^3(-1) = -e^3\]

  53. zepdrix
    • one year ago
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    You and your ways -_- lol You gotta remember your trig stuff silly billy! \(\large\rm \cos(\pi)=-1\) <- yes. I don't like the 3.14 though lol

  54. anonymous
    • one year ago
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    Lol sorry! But thank you soo much!! You are really good at helping others! :D

  55. zepdrix
    • one year ago
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    yay team \c:/ we did it!

  56. anonymous
    • one year ago
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    Thanks @Mikeyy1992! You were very helpful too! :)

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