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abannavong

  • 2 years ago

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

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  1. hartnn
    • 2 years ago
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    multiply and divide by x , to get the function in the form of sin x/ x

  2. abannavong
    • 2 years ago
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    can u draw it out so i can see what ur talking about

  3. hartnn
    • 2 years ago
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    \(\huge\frac{sin x}{x}\:\frac{x}{\sqrt[3]{x}}\)

  4. abannavong
    • 2 years ago
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    oh ok

  5. abannavong
    • 2 years ago
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    and then from there which method would u use

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

  7. abannavong
    • 2 years ago
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    oh ok! omg now it makes sense thank you @hartnn

  8. hartnn
    • 2 years ago
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    welcome :) what did u get the final answer as ?

  9. abannavong
    • 2 years ago
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    hang on lolz

  10. abannavong
    • 2 years ago
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    is it 0?

  11. hartnn
    • 2 years ago
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    yup, its 0 good work :)

  12. abannavong
    • 2 years ago
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    yay!!!

  13. abannavong
    • 2 years ago
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    thank you!!

  14. abannavong
    • 2 years ago
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    i need help again actually with number 84 now

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  15. hartnn
    • 2 years ago
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    ok, if f(x) = root x what will be f(x+h) ?? u know?

  16. abannavong
    • 2 years ago
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    im confused on tht part

  17. hartnn
    • 2 years ago
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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?

  18. abannavong
    • 2 years ago
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    ok

  19. abannavong
    • 2 years ago
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    and then what?

  20. hartnn
    • 2 years ago
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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 ?

  21. abannavong
    • 2 years ago
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    yeah i got tht part

  22. hartnn
    • 2 years ago
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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 ?

  23. abannavong
    • 2 years ago
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    oh so ur multiplying it by its conjugate right ?

  24. hartnn
    • 2 years ago
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    yup, so that i get only h in numerator and that i can cancel it with h in denominator ok?

  25. abannavong
    • 2 years ago
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    oh ok

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

  27. abannavong
    • 2 years ago
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    wait a minute now im confused a bit

  28. hartnn
    • 2 years ago
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    because itsf(x+h) - f(x) which is \(\sqrt{x+h} \) \(-\sqrt{x} \)

  29. abannavong
    • 2 years ago
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    oh ok

  30. hartnn
    • 2 years ago
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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 ?

  31. abannavong
    • 2 years ago
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    would it still be h in the numerator i think..

  32. hartnn
    • 2 years ago
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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?

  33. abannavong
    • 2 years ago
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    do you have \[\sqrt{x+h} + \sqrt{x}\]

  34. abannavong
    • 2 years ago
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    right?

  35. abannavong
    • 2 years ago
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    idk lolz

  36. hartnn
    • 2 years ago
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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

  37. abannavong
    • 2 years ago
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    \[\sqrt{x+0} +\sqrt{x}\]

  38. abannavong
    • 2 years ago
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    is it?

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

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

  41. hartnn
    • 2 years ago
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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 ?

  42. abannavong
    • 2 years ago
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    oh ok

  43. abannavong
    • 2 years ago
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    thanks!

  44. hartnn
    • 2 years ago
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    welcome :)

  45. abannavong
    • 2 years ago
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    u r a very helpful math teacher!

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