Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mathcalculus

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

  • 8 months ago
  • 8 months ago

  • This Question is Closed
  1. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    • 8 months ago
  2. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    when we evaluate the limit, do we not substitute?

    • 8 months ago
  3. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    this is where i'm confused.

    • 8 months ago
  4. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    in this case, i got the answer right. i kind of took a smart guess

    • 8 months ago
  5. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

    • 8 months ago
  6. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

    When the limit is NOT giving you an indeterminate form, yes you're allowed to substitute the value directly in :)

    • 8 months ago
  7. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @pgpilot326 a little

    • 8 months ago
  8. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    indeterminate form meaning.... 0/0?

    • 8 months ago
  9. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 8 months ago
  10. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i got the numerator..319/8 and the denominator -77/8

    • 8 months ago
  11. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    And indeterminate form can mean more than just 0/0. For the purposes of calc 1, yeah, 0/0

    • 8 months ago
  12. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i simply subtracted and simplified.

    • 8 months ago
  13. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    but i saw that they asked infinity or non infinity etc etc

    • 8 months ago
  14. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    so i guessed.

    • 8 months ago
  15. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix

    • 8 months ago
  16. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 8 months ago
  17. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    oh...k

    • 8 months ago
  18. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    so what i did was wrong?

    • 8 months ago
  19. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 8 months ago
  20. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    so.... how should i do it?

    • 8 months ago
  21. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 8 months ago
  22. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    im more confused now. :/

    • 8 months ago
  23. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 8 months ago
  24. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah what if i wanted to use 11/8

    • 8 months ago
  25. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    it's kidn of the same thing... right?

    • 8 months ago
  26. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i see that one goes really high up and the other goes really down.

    • 8 months ago
  27. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    Then you will get an undefined answer. An undefined answer does you no good.

    • 8 months ago
  28. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    no i didn't get an undefined answer...

    • 8 months ago
  29. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I apologize, I did not look correctly. I'm brain cramping then, ignore what I've said x_x

    • 8 months ago
  30. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    oh boy.

    • 8 months ago
  31. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    x\

    • 8 months ago
  32. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 8 months ago
  33. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait, what? You DIDNT get an undefined? 11- 8(11/8)?

    • 8 months ago
  34. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • 8 months ago
  35. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @pgpilot326 show it pls

    • 8 months ago
  36. Psymon
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah. nvm, I'm clearly being confusing. I'll leave, sorry.

    • 8 months ago
  37. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    #1) can i simply substitute 11/8 into the equation? yes or no?

    • 8 months ago
  38. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    okay no, this method confused me from the beginning. :/

    • 8 months ago
  39. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix wait what?

    • 8 months ago
  40. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

    lemme erase all that nonsense and see if i can explain this :)

    • 8 months ago
  41. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large \lim_{x\to11/8^+} \frac{-29x}{11-8x} \qquad\to\qquad \frac{39.875}{0}\] 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?

    • 8 months ago
  42. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @zepdrix how'd you get that number?

    • 8 months ago
  43. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

    I plugged 11/8 directly in.

    • 8 months ago
  44. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    negative?

    • 8 months ago
  45. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok i see where you got the number from

    • 8 months ago
  46. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

    negative? Is that in response to the question about the denominator?

    • 8 months ago
  47. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i think, not sure. i think it's negative.

    • 8 months ago
  48. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • 8 months ago
  49. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • 8 months ago
  50. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 8 months ago
  51. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    But don't forget what the top is doing too.

    • 8 months ago
  52. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • 8 months ago
  53. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i know this: ex: 1/x all real numbers except x=0

    • 8 months ago
  54. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    with this problem. im just so so confused on to how to approach it.

    • 8 months ago
  55. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    ex: step by step. and the understanding part is the hardest part.

    • 8 months ago
  56. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\lim_{x \rightarrow a}f \left( x \right)=c \iff \lim_{x \rightarrow a ^{+}}f \left( x \right)=\lim_{x \rightarrow a ^{-}}f \left( x \right)=c\] 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.

    • 8 months ago
  57. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    lol it does.

    • 8 months ago
  58. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks i understand it much better.

    • 8 months ago
  59. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i see that the left side is coming from the negative side. and right from the positive.

    • 8 months ago
  60. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @pgpilot326

    • 8 months ago
  61. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    now let's say for this problem:

    • 8 months ago
  62. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

    • 8 months ago
  63. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    h/o

    • 8 months ago
  64. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i substituted all 7 in.... the problem is the signs right and left.

    • 8 months ago
  65. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i got them wrong.

    • 8 months ago
  66. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    i typed in n, p, dne

    • 8 months ago
  67. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    @pgpilot326

    • 8 months ago
  68. mathcalculus
    Best Response
    You've already chosen the best response.
    Medals 0

    ?

    • 8 months ago
  69. pgpilot326
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[\lim_{x \rightarrow 7}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=-\infty \iff \lim_{x \rightarrow 7^{-}}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=\lim_{x \rightarrow 7^{+}}\left( \frac{ -2x+1 }{ \left( x-7 \right)^{2} } \right)=-\infty\]

    • 8 months ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.