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mathcalculus

  • one year ago

Evaluate the limit: can someone explain me this? (attached below)

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  1. mathcalculus
    • one year ago
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  2. mathcalculus
    • one year ago
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    when we evaluate the limit, do we not substitute?

  3. mathcalculus
    • one year ago
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    this is where i'm confused.

  4. mathcalculus
    • one year ago
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    in this case, i got the answer right. i kind of took a smart guess

  5. pgpilot326
    • one year ago
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    at x=11/8 the bottom goes to 0, right? But coming from the right, the bottom is neg, as is the top so it goes to inf. Coming from the left, the bottom is pos and the top is neg so it goes to minf. Make sense?

  6. zepdrix
    • one year ago
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    When the limit is NOT giving you an indeterminate form, yes you're allowed to substitute the value directly in :)

  7. mathcalculus
    • one year ago
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    @pgpilot326 a little

  8. mathcalculus
    • one year ago
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    indeterminate form meaning.... 0/0?

  9. Psymon
    • one year ago
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    You're really just testing points on each side of your undefined point. You need to see what the behavior is on each side, so pick a point on each side and see what it's doing.

  10. mathcalculus
    • one year ago
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    i got the numerator..319/8 and the denominator -77/8

  11. Psymon
    • one year ago
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    And indeterminate form can mean more than just 0/0. For the purposes of calc 1, yeah, 0/0

  12. mathcalculus
    • one year ago
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    i simply subtracted and simplified.

  13. mathcalculus
    • one year ago
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    but i saw that they asked infinity or non infinity etc etc

  14. mathcalculus
    • one year ago
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    so i guessed.

  15. mathcalculus
    • one year ago
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    @zepdrix

  16. Psymon
    • one year ago
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    Yeah, because it will go to infinity or negative infinity. You're approaching an asymptote, not an actual point. The graph will just shoot up or down. But the way to see whether it shoots up or down is to test points on each side of the asymptote.

  17. mathcalculus
    • one year ago
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    oh...k

  18. mathcalculus
    • one year ago
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    so what i did was wrong?

  19. Psymon
    • one year ago
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    Kinda, yeah. I would just pick a point really close to the asymptote on each side. If it's really close, your number will either be really high in the positive direction or really high in the negative direction.

  20. mathcalculus
    • one year ago
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    so.... how should i do it?

  21. Psymon
    • one year ago
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    Like I picked the number 5/4 for the left side and got -145/4. Picking 3/2 for the right side I got 87/2. You can kinda see how one gets high fast and one goes low fast.

  22. mathcalculus
    • one year ago
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    im more confused now. :/

  23. Psymon
    • one year ago
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    I would just personally pick numbers on each side of the asymptote that are very close to it. If they're close enough, the answer should tell you pretty clearly if you're going way negative or way positive. Maybe the others can explain better x_x

  24. mathcalculus
    • one year ago
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    yeah what if i wanted to use 11/8

  25. mathcalculus
    • one year ago
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    it's kidn of the same thing... right?

  26. mathcalculus
    • one year ago
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    i see that one goes really high up and the other goes really down.

  27. Psymon
    • one year ago
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    Then you will get an undefined answer. An undefined answer does you no good.

  28. mathcalculus
    • one year ago
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    no i didn't get an undefined answer...

  29. Psymon
    • one year ago
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    Oh, I apologize, I did not look correctly. I'm brain cramping then, ignore what I've said x_x

  30. mathcalculus
    • one year ago
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    oh boy.

  31. mathcalculus
    • one year ago
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    x\

  32. pgpilot326
    • one year ago
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    look, when x is near 11/8 but more, 11 - 8x is gonna be negative. neg over neg is pos. when x is near 11/8 but less, 11 - 8x is gonna be pos. neg over pos is neg. In both cases, 11 - 8x is headed towards 0 as x goes to 11/8.

  33. Psymon
    • one year ago
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    Wait, what? You DIDNT get an undefined? 11- 8(11/8)?

  34. Psymon
    • one year ago
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    I guess I was right and confused myself -_- Well, my advice would to just do what I tried to explain. If you can test points on both sides of the asymptote and see theyre really positive or really negative, that you should tell you your answer. As to how you didn't get an undefined when you plugged in 11/8, I'm not sure?

  35. mathcalculus
    • one year ago
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    @pgpilot326 show it pls

  36. Psymon
    • one year ago
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    Yeah. nvm, I'm clearly being confusing. I'll leave, sorry.

  37. mathcalculus
    • one year ago
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    #1) can i simply substitute 11/8 into the equation? yes or no?

  38. mathcalculus
    • one year ago
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    okay no, this method confused me from the beginning. :/

  39. mathcalculus
    • one year ago
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    @zepdrix wait what?

  40. zepdrix
    • one year ago
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    lemme erase all that nonsense and see if i can explain this :)

  41. zepdrix
    • one year ago
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    \[\large \lim_{x\to11/8^+} \frac{-29x}{11-8x} \qquad\to\qquad \frac{39.875}{0}\] Plugging the fraction directly in shows us that we're approaching this form. Since we're approaching a zero in the denominator, it means there's an asymptote at x=11/8. We need to figure something out though. Are we approaching positive or negative infinity? Well our numerator is negative. How about our denominator? When x is a tiny bit bigger than 11/8, is the denominator negative or positive?

  42. mathcalculus
    • one year ago
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    @zepdrix how'd you get that number?

  43. zepdrix
    • one year ago
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    I plugged 11/8 directly in.

  44. mathcalculus
    • one year ago
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    negative?

  45. mathcalculus
    • one year ago
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    oh ok i see where you got the number from

  46. zepdrix
    • one year ago
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    negative? Is that in response to the question about the denominator?

  47. mathcalculus
    • one year ago
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    i think, not sure. i think it's negative.

  48. zepdrix
    • one year ago
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    Yes, very good! Because \(\large -8x\) is slightly larger than \(\large 11\) when we plug in a value larger than 11/8. So \(\large 11-8x\) is giving us a negative value when we approach from the larger side.

  49. pgpilot326
    • one year ago
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    think of the function 1/x. when x goes to zero, the function goes to inf from the right because x is positive and it goes to negative infinite from the left because x is negative. the limit doesn't exist because it goes to 2 different places when coming from the left and right. You're problem is the same except it goes to 0 on the bottom when x =11/8. So the limit won't exist at x=11/8 if the limit goes to 2 different places coming from the left and right. 11-8x=0 when x=11/8. But if x>11/8, 11-8x is negative and if x<11/8, 11-8x is positive. This is how you determine where it's heading as you approach the zero from either side.

  50. zepdrix
    • one year ago
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    ^ yah make sure you're comfortable with \(\large \lim_{x\to0}\dfrac{1}{x}\) as pilot explained.

  51. pgpilot326
    • one year ago
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    But don't forget what the top is doing too.

  52. mathcalculus
    • one year ago
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    okay, so what are the exact rules to every limit problem. keeping in mind y/x if denominator is 0 then it's negative?

  53. mathcalculus
    • one year ago
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    i know this: ex: 1/x all real numbers except x=0

  54. mathcalculus
    • one year ago
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    with this problem. im just so so confused on to how to approach it.

  55. mathcalculus
    • one year ago
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    ex: step by step. and the understanding part is the hardest part.

  56. pgpilot326
    • one year ago
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    \[\lim_{x \rightarrow a}f \left( x \right)=c \iff \lim_{x \rightarrow a ^{+}}f \left( x \right)=\lim_{x \rightarrow a ^{-}}f \left( x \right)=c\] This says the limit exists if and only if the limit from the left and right are the same. You can have different left and right limits, in which case the overall limit will not exist. For example, let \[f \left( x \right)=\frac{ 1 }{ \left( x-3 \right) }\] in order to determine \[\lim_{x \rightarrow 3}f \left( x \right)\] we need to look at what happens on the left and right as x approaches 3. \[\lim_{x \rightarrow 3^{+}}\frac{ 1 }{ \left( x-3 \right) }=+\infty\]because \[\left( x-3 \right)>0, \forall x>3\]. Likewise, \[\lim_{x \rightarrow 3^{-}}\frac{ 1 }{ \left( x-3 \right) }=-\infty\]because \[\left( x-3 \right)<0, \forall x<3\]. The bottom is going to 0 as x gets closer and closer to 3. But coming from above 3 (like 3.1, 3.01, 3.001), the bottom is still positive as it goes to zero. yetcoming from below 3 (like 2.9, 2.99, 2.999) the bottom is still negative as it goes to zero. I hope that helps cause it was alot to type.

  57. mathcalculus
    • one year ago
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    lol it does.

  58. mathcalculus
    • one year ago
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    thanks i understand it much better.

  59. mathcalculus
    • one year ago
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    i see that the left side is coming from the negative side. and right from the positive.

  60. mathcalculus
    • one year ago
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    @pgpilot326

  61. mathcalculus
    • one year ago
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    now let's say for this problem:

  62. pgpilot326
    • one year ago
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    For all of these evaluate the top and bottom separately and them bring them together. the bottom is always positive, no matter what side you come from. That's because you're squaring the bottom. The top is always going to be negative. Thus all of them will be negative infinity. Do you see?

  63. mathcalculus
    • one year ago
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    h/o

  64. mathcalculus
    • one year ago
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    i substituted all 7 in.... the problem is the signs right and left.

  65. mathcalculus
    • one year ago
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    i got them wrong.

  66. mathcalculus
    • one year ago
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    i typed in n, p, dne

  67. mathcalculus
    • one year ago
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    @pgpilot326

  68. mathcalculus
    • one year ago
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    ?

  69. pgpilot326
    • one year ago
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    You can break up the limit into the limit of the top times the limit of the bottom. If the limit exists in each case, then all of the limits (left, right and overall) will be the same. \[\lim_{x \rightarrow 7}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=\lim_{x \rightarrow 7}\left( -2x+1 \right) \times \lim_{x \rightarrow 7}\left( \frac{ 1 }{ \left( x-7 \right)^{2} } \right)\] \[\lim_{x \rightarrow 7}-2x+1 = -13 \iff \lim_{x \rightarrow 7^{-}}-2x+1 =\lim_{x \rightarrow 7^{+}}-2x+1 = -13\] Also \[\lim_{x \rightarrow 7}\left( \frac{ 1 }{ \left( x-7 \right)^{2} } \right)=+\infty \iff \lim_{x \rightarrow 7^{-}}\left( \frac{ 1 }{ \left( x-7 \right)^{2} } \right)=\lim_{x \rightarrow 7^{+}}\left( \frac{ 1 }{ \left( x-7 \right)^{2} } \right)=+\infty\] So \[\lim_{x \rightarrow 7}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=-\infty \iff \lim_{x \rightarrow 7^{-}}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=\lim_{x \rightarrow 7^{+}}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=-\infty\]

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