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anonymous

  • 4 years ago

lim t->+oo (t(eˆtˆ-3 - 1)/ sin(tˆ-1)

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  1. anonymous
    • 4 years ago
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    please help me

  2. dumbcow
    • 4 years ago
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    \[\lim_{? \rightarrow ?} \frac{t(e^{1/t^{3}}-1}{\sin(1/t)}\] is that right?

  3. anonymous
    • 4 years ago
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    lim t-> +oo

  4. dumbcow
    • 4 years ago
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    ok

  5. anonymous
    • 4 years ago
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    t*(eˆtˆ-3)-1

  6. dumbcow
    • 4 years ago
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    looks like you get 0/0 use L'hopitals rule and take derivative of top and bottom

  7. dumbcow
    • 4 years ago
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    so the minus 1 is on outside ?

  8. anonymous
    • 4 years ago
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    oh no in inside

  9. anonymous
    • 4 years ago
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    Let x = 1 / t, hence when t -> oo, x -> 0 hence the limit becomes Lim x -> 0 (e^x^3 - 1) / (sinx /x)

  10. anonymous
    • 4 years ago
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    lim x -> 0 sinx / x = 1 and the top becomes 0, hence the answer is 0

  11. anonymous
    • 4 years ago
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    did you get it?

  12. anonymous
    • 4 years ago
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    but the result of this limit is 0?

  13. anonymous
    • 4 years ago
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    yeah, that is acceptable

  14. anonymous
    • 4 years ago
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    it is a finite number

  15. anonymous
    • 4 years ago
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    did you get the process of reaching the answer

  16. anonymous
    • 4 years ago
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    aahhh ok

  17. anonymous
    • 4 years ago
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    you didi the substitution right? (sorry for my english!)

  18. anonymous
    • 4 years ago
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    yeah I did, I made x = 1 / t, so if t -> oo then x should be -> 0

  19. dumbcow
    • 4 years ago
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    mainaknag so t = 1/x right i don;t think you substituted correctly --> (e^x3 -1)/x*sinx

  20. anonymous
    • 4 years ago
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    I'm confuse

  21. anonymous
    • 4 years ago
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    yeah I am sorry

  22. anonymous
    • 4 years ago
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    ok let me try again, thank you dumbcow

  23. dumbcow
    • 4 years ago
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    but you are right, the limit is zero though :)

  24. anonymous
    • 4 years ago
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    hmm, I think so

  25. dumbcow
    • 4 years ago
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    not sure how to show it though...L'hopitals rule gets me nowhere

  26. anonymous
    • 4 years ago
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    you need to divide by x3 both numerator and denominator

  27. anonymous
    • 4 years ago
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    lim y -> 0 (e^y - 1) / y = 1, where y = x3 since x -> 0, y -> 0

  28. anonymous
    • 4 years ago
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    wait coming with full solution

  29. anonymous
    • 4 years ago
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    \[\lim_{t \rightarrow \infty} t (e ^{1/t ^{3}} - 1) / \sin (1 / t)\] Let x -> 1 / t, then x -> 0 And the limit becomes -\[\lim_{x \rightarrow 0} (e ^{x ^{3}} - 1) / x.sinx = \lim_{x \rightarrow 0} \lim_{y \rightarrow 0}(1 /y).(e ^{y} - 1) / (1 / x) . (x / x) . (Sinx / x)\] the numerator -> 1, lim x -> 0 sinx / x = 1, lim x -> 0 x / x = 1 Hence the limit becomes \[\lim_{x \rightarrow 0} 1 / (1 / x) = \lim_{x \rightarrow 0} x = 0\]

  30. anonymous
    • 4 years ago
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    just need to mention I substituted y = x3

  31. anonymous
    • 4 years ago
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    thank you

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