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Jamierox4ev3r

  • one year ago

Check my work please. Not sure if I'm going in the right direction here

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  1. Jamierox4ev3r
    • one year ago
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    \[\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\]

  2. Jamierox4ev3r
    • one year ago
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    I assume that i would have to multiply both numerator and denominator with a conjugate

  3. anonymous
    • one year ago
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    correct

  4. Jamierox4ev3r
    • one year ago
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    and that conjugate would be \(\large\sqrt{9+h}+3\)

  5. Jamierox4ev3r
    • one year ago
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    alright. At least that assumption was correct

  6. Jamierox4ev3r
    • one year ago
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    so from there, since i would multiply \(\sqrt{9+h}-3\times \sqrt{9+h}+3\)

  7. Jamierox4ev3r
    • one year ago
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    since \(\sqrt{x^{2}}\) is equal to \(x\), then \(\sqrt{9+h}\times \sqrt{9+h}\) would be equal to \(9+h\)

  8. Jamierox4ev3r
    • one year ago
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    Since the middle terms cancel out, I believe we are left with \(\Large\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{9+h} +3}\)

  9. anonymous
    • one year ago
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    in the numerator 9+h-9 in the denominator ? you are correct.

  10. Jamierox4ev3r
    • one year ago
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    Nice! So I wasn't sure if I was multiplying things correctly oh, and I kind of forgot to close the parentheses in the denominator, but other that, it looks good to you? @surjithayer

  11. Jamierox4ev3r
    • one year ago
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    \(\Large\lim_{h \rightarrow 0} \frac{\color{red}h}{\color{red}h(\sqrt{9+h} +3}\) The things in red will cancel out, so you should be left with \(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\)

  12. Jamierox4ev3r
    • one year ago
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    @nincompoop from here, would I just plug in zero in place of h? (assuming, of course, that I hadn't taken any erroneous steps)

  13. AravindG
    • one year ago
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    I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.

  14. Jamierox4ev3r
    • one year ago
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    at this point, it doesn't.

  15. Jamierox4ev3r
    • one year ago
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    took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3

  16. nincompoop
    • one year ago
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    assuming that you've done correctly up to the last portion, all you have to do is plug in the value of h = 0 and simplify

  17. nincompoop
    • one year ago
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    you need to show me the original function tho

  18. Jamierox4ev3r
    • one year ago
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    alright. I've already verified that most of my steps were correct with surji, and here is the original fxn \(\Large\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\)

  19. Jamierox4ev3r
    • one year ago
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    @AravindG I think I already outlined that somewhere up there :P

  20. Jhannybean
    • one year ago
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    You got this :) You just need to stay confident!

  21. Jamierox4ev3r
    • one year ago
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    thanks :) @Jhannybean

  22. AravindG
    • one year ago
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    Yeah you are right. The limit exists.

  23. Jhannybean
    • one year ago
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    \[\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right) = \frac{1}{6}}\]

  24. nincompoop
    • one year ago
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    there are few key points that you probably already learned whether or not limit may or may not exist based on the function itself without doing all of the work. it cuts many of your work

  25. nincompoop
    • one year ago
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    thanks, jhan

  26. nincompoop
    • one year ago
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    I knew there was something odd LAUGHING OUT LOUD

  27. nincompoop
    • one year ago
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    jhan :*

  28. Jamierox4ev3r
    • one year ago
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    Alright. Good to know. \(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\) \(\Large\frac{1}{\sqrt{9+0} +3} = \frac{1}{3+3}\)

  29. Jhannybean
    • one year ago
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    Yep.

  30. Jamierox4ev3r
    • one year ago
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    and that would give you \(\Large\frac{1}{6}\). Alright then, that makes sense!

  31. Jamierox4ev3r
    • one year ago
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    So at that point, you can plug in 0, since it wouldn't become undefined. g'nice

  32. nincompoop
    • one year ago
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    ye why is that?

  33. nincompoop
    • one year ago
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    why can't we just plug it in immediately?

  34. Jamierox4ev3r
    • one year ago
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    because then the whole fraction would be set over 0. As long as the denominator is zero, we conclude that the limit does not exist, when in reality it does

  35. nincompoop
    • one year ago
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    interesting

  36. Jamierox4ev3r
    • one year ago
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    At least I think that's what the prof. said o-o lol

  37. nincompoop
    • one year ago
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    so what does it mean that the limit = 1/6 ?

  38. Jamierox4ev3r
    • one year ago
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    why dost thou question me profusely? XD

  39. nincompoop
    • one year ago
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    fundamental concepts and analysis

  40. Jamierox4ev3r
    • one year ago
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    I see o-o

  41. Jhannybean
    • one year ago
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    That means at some value of x, the function will approach y at 1/6,

  42. Jamierox4ev3r
    • one year ago
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    ^that's the technical meaning of a limit

  43. nincompoop
    • one year ago
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    and then what?

  44. Jamierox4ev3r
    • one year ago
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    At this point, I'm not sure what you're asking

  45. nincompoop
    • one year ago
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    haha

  46. Jamierox4ev3r
    • one year ago
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    Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? o-o

  47. nincompoop
    • one year ago
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    just telling me what the formal definition of limit does not tell me what you understand about it, that's all

  48. Jamierox4ev3r
    • one year ago
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    Something to do with reimann sum right? :P lol

  49. Jamierox4ev3r
    • one year ago
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    Anyhow, thank you, I'll close this for now :)

  50. Jamierox4ev3r
    • one year ago
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    @nincompoop Is the reimann sum the same as the epsilon-delta limit definition?! Because if it is, then we barely went over that in class. And if they are indeed the same thing, then I do understand what you're saying

  51. nincompoop
    • one year ago
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    they are related

  52. Jamierox4ev3r
    • one year ago
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    I see

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