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anonymous

  • 4 years ago

given that f(x) =(∛x-1)/(x-1) then its limit as x approaches 1 is equal to? a.0 b.1 c.-1 d.does not exist

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  1. anonymous
    • 4 years ago
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    is (x-1) all under the cube root?

  2. anonymous
    • 4 years ago
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    No option seem correct

  3. anonymous
    • 4 years ago
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    no

  4. anonymous
    • 4 years ago
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    x-1 not under the cube root

  5. myininaya
    • 4 years ago
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    \[\text{ let } u=x^\frac{1}{3} ; x->1 => u->(1)^\frac{1}{3}=1\] so we have \[\lim_{u \rightarrow 1}\frac{u-1}{u^3-1}=\lim_{u \rightarrow 1}\frac{u-1}{(u-1)(u^2+u+1)}\]

  6. myininaya
    • 4 years ago
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    that should help out

  7. nenadmatematika
    • 4 years ago
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    solution is 1/3 if you wrote it correctly

  8. anonymous
    • 4 years ago
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    so if theres no answer i put E.

  9. nenadmatematika
    • 4 years ago
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    I suggest you type it again more carefuly and post again...so I would be sure

  10. anonymous
    • 4 years ago
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    It's 0, look at myininaya's solution... \[\frac{1-1}{1^2+1+1}=\frac{0}{3}=0\]

  11. myininaya
    • 4 years ago
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    hey rick that top part cancels with (x-1) though

  12. myininaya
    • 4 years ago
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    i mean u-1

  13. anonymous
    • 4 years ago
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    ah right, my mistake

  14. myininaya
    • 4 years ago
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    so you do get 1/3

  15. nenadmatematika
    • 4 years ago
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    Nenad is the champ :D

  16. nenadmatematika
    • 4 years ago
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    hahaha...kidding

  17. anonymous
    • 4 years ago
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    lol

  18. anonymous
    • 4 years ago
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    this one has a sqr (∛x-1) the other 1 hast no cube root (x-1) and i feel nenad is the champ XD + myiniaya XD

  19. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 1}\frac{\sqrt[3]{x-1}}{x-1}\] are you saying this is the problem?

  20. anonymous
    • 4 years ago
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    that's what i asked him earlier..he said no

  21. anonymous
    • 4 years ago
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    YES THATS the one thanks

  22. anonymous
    • 4 years ago
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    sorry rick im confused

  23. anonymous
    • 4 years ago
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    so what is my answer d or e? e for error

  24. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 1}\frac{\sqrt[3]{x-1} \cdot (x-1)^3}{(x-1) \cdot (x-1)^3}\] \[\lim_{x \rightarrow 1}\frac{(x-1)}{(x-1)(x-1)^3}=\lim_{x \rightarrow 1}\frac{1}{(x-1)^4}\] I get the limit does not exist

  25. nenadmatematika
    • 4 years ago
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    does not exist

  26. anonymous
    • 4 years ago
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    or \[\infty\]

  27. anonymous
    • 4 years ago
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    thanks rick

  28. anonymous
    • 4 years ago
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    d is wrong to i got the result of my question i got 16/20 =90 ^^ thanks to all of u

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