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anonymous

  • 5 years ago

Evaluate the following limit: \[sqrt{(9+11x^{2})} / (6+3x)]\

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  1. anonymous
    • 5 years ago
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    as x -> inf and as x-> -(inf)

  2. anonymous
    • 5 years ago
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    apply La Hopitals rule

  3. anonymous
    • 5 years ago
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    differentiate top, differentiate the bottom, and try again

  4. anonymous
    • 5 years ago
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    y= ( 9 + 11x^2)^1/2 dy/dx = (1/2) ( 9 + 11x^2 )^-1/2 * 22x

  5. anonymous
    • 5 years ago
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    = 11x / ( sqrt ( 9+11x^2) )

  6. anonymous
    • 5 years ago
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    so we want to take the limit of\[\frac{ \frac{11x}{\sqrt{9+11x^2} } } { 3}\]

  7. anonymous
    • 5 years ago
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    yea, that's what i got so far

  8. anonymous
    • 5 years ago
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    applying the rule again because it is still indeterminate form "infinity/infinity"

  9. anonymous
    • 5 years ago
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    so the (1/3) factor , bring that out then we have 11 / [ 11x / sqrt(9+11x^2) ]

  10. anonymous
    • 5 years ago
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    so that gives what sqrt(9+11x^2) / x I think

  11. anonymous
    • 5 years ago
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    \[\frac{\sqrt{9+11x^2}}{x}\]

  12. anonymous
    • 5 years ago
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    sstill indeterminate , but if we apply the rule once more that x on the denominator will vanish

  13. anonymous
    • 5 years ago
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    yeh I think something possibly went wrong, because it we differentiate that then we get back to 11x/ sqrt (9+11x^2)

  14. anonymous
    • 5 years ago
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    I would just go with dominate term analysis here I think

  15. anonymous
    • 5 years ago
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    I would just go with dominate term analysis here I think

  16. anonymous
    • 5 years ago
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    when i apply the lospitl rule i don't have to do a quotien rule between the two func tion right?

  17. anonymous
    • 5 years ago
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    i got the answers right thou thanks to inf is sqrt11/3 and to -(inf) is -sqrt11/3

  18. anonymous
    • 5 years ago
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    dominate term up the top is \[x \sqrt{11}\] the dominate term on the bottom is x ( obviously )

  19. anonymous
    • 5 years ago
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    so the quotient of these gives the answers sqrt(11) , and then you remember the factor of (1/3) we had earlier

  20. anonymous
    • 5 years ago
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    how is it \[x sqrt {11}\]

  21. anonymous
    • 5 years ago
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    do i just forget about the 9 ?

  22. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow \infty}\frac{\sqrt{9+11x^{2}}}{6+3x}\] \[ = \lim_{x \rightarrow \infty} \frac{x\sqrt{\frac{9}{x^2}+11}}{x(\frac{6}{x}+3)}\]

  23. anonymous
    • 5 years ago
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    Becomes pretty easy from there.

  24. anonymous
    • 5 years ago
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    Looks like \[\frac{\sqrt{11}}{3}\]

  25. anonymous
    • 5 years ago
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    Did you follow that?

  26. anonymous
    • 5 years ago
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    i didn't get how yu got sqrt11/3 from that. Did you just ignore everything that had an x ?

  27. anonymous
    • 5 years ago
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    No. I took the limit as x goes to infinity. \(9/x^2\) will go to 0. 6/x will go to 0. The only things left will be the sqrt{11} and the 3 in the denominator.

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