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abannavong Group Title

can someone help my find the limit if x->0 sinx/ cubed root of x

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    multiply and divide by x , to get the function in the form of sin x/ x

    • one year ago
  2. abannavong Group Title
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    can u draw it out so i can see what ur talking about

    • one year ago
  3. hartnn Group Title
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    \(\huge\frac{sin x}{x}\:\frac{x}{\sqrt[3]{x}}\)

    • one year ago
  4. abannavong Group Title
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    oh ok

    • one year ago
  5. abannavong Group Title
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    and then from there which method would u use

    • one year ago
  6. hartnn Group Title
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    u know the formula : \[\lim_{x \rightarrow 0}\: sinx / x = 1\] ? use that for first fraction.

    • one year ago
  7. abannavong Group Title
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    oh ok! omg now it makes sense thank you @hartnn

    • one year ago
  8. hartnn Group Title
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    welcome :) what did u get the final answer as ?

    • one year ago
  9. abannavong Group Title
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    hang on lolz

    • one year ago
  10. abannavong Group Title
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    is it 0?

    • one year ago
  11. hartnn Group Title
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    yup, its 0 good work :)

    • one year ago
  12. abannavong Group Title
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    yay!!!

    • one year ago
  13. abannavong Group Title
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    thank you!!

    • one year ago
  14. abannavong Group Title
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    i need help again actually with number 84 now

    • one year ago
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  15. hartnn Group Title
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    ok, if f(x) = root x what will be f(x+h) ?? u know?

    • one year ago
  16. abannavong Group Title
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    im confused on tht part

    • one year ago
  17. hartnn Group Title
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    ok, to get f(x+h) just replace x by (x+h) in f(x) so f(x+h) will be \(\sqrt{x+h}\) ok?

    • one year ago
  18. abannavong Group Title
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    ok

    • one year ago
  19. abannavong Group Title
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    and then what?

    • one year ago
  20. hartnn Group Title
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    i m using h instead of delta x just for convenience here. so now u put that in {f(x+h)-f(x) }/ h to get \(\frac{\sqrt{x+h}-\sqrt{h}}{h}\) got this ?

    • one year ago
  21. abannavong Group Title
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    yeah i got tht part

    • one year ago
  22. hartnn Group Title
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    now mutiply and divide by \(\sqrt{x+h}+\sqrt{h}\) and use (a+b)(a-b) = a^2 - b^2 and tell me what u get in numerator ?

    • one year ago
  23. abannavong Group Title
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    oh so ur multiplying it by its conjugate right ?

    • one year ago
  24. hartnn Group Title
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    yup, so that i get only h in numerator and that i can cancel it with h in denominator ok?

    • one year ago
  25. abannavong Group Title
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    oh ok

    • one year ago
  26. hartnn Group Title
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    i made a typo ..... it should be \(-\sqrt{x} \) and not \(-\sqrt{h} \) in the numerator

    • one year ago
  27. abannavong Group Title
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    wait a minute now im confused a bit

    • one year ago
  28. hartnn Group Title
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    because itsf(x+h) - f(x) which is \(\sqrt{x+h} \) \(-\sqrt{x} \)

    • one year ago
  29. abannavong Group Title
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    oh ok

    • one year ago
  30. hartnn Group Title
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    so now now mutiply and divide by \(\sqrt{x+h}+\sqrt{x}\) and use (a+b)(a-b) = a^2 - b^2 and tell me what u get in numerator ?

    • one year ago
  31. abannavong Group Title
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    would it still be h in the numerator i think..

    • one year ago
  32. hartnn Group Title
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    yup, because (x+h)- x = x+h-x = h this h cancels out with h in denominator and u have only 1 in numerator, right ? and what u have in denominator now?

    • one year ago
  33. abannavong Group Title
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    do you have \[\sqrt{x+h} + \sqrt{x}\]

    • one year ago
  34. abannavong Group Title
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    right?

    • one year ago
  35. abannavong Group Title
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    idk lolz

    • one year ago
  36. hartnn Group Title
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    yes, u are right, be confident :) now just put h =0 because lim h-> 0 and tell me what u get in denominator after putting h=0

    • one year ago
  37. abannavong Group Title
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    \[\sqrt{x+0} +\sqrt{x}\]

    • one year ago
  38. abannavong Group Title
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    is it?

    • one year ago
  39. hartnn Group Title
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    yup, thats your denominator = \(2\sqrt{x}\) and then your final answer would be \(\huge\frac{1}{2\sqrt{x}}\)

    • one year ago
  40. abannavong Group Title
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    hang on im confused how do u get \[\frac{ 1 }{ 2\sqrt{x} }\]

    • one year ago
  41. hartnn Group Title
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    \(\sqrt{x+0} +\sqrt{x}=\sqrt{x} +\sqrt{x}=2\sqrt{x}\) and there was 1 in the numerator, previously, right ?

    • one year ago
  42. abannavong Group Title
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    oh ok

    • one year ago
  43. abannavong Group Title
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    thanks!

    • one year ago
  44. hartnn Group Title
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    welcome :)

    • one year ago
  45. abannavong Group Title
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    u r a very helpful math teacher!

    • one year ago
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