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anonymous

  • 5 years ago

use L'Hopital's rule to calculate the following derivatives ; lim x->1 x^1/1-x lim x-> x^2-e^x/x

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  1. anonymous
    • 5 years ago
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    the answer for first is 0 . just need the steps.

  2. anonymous
    • 5 years ago
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    you mean the following limits?

  3. anonymous
    • 5 years ago
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    is the first one \[\lim_{x \rightarrow 1}{x \over 1-x}?\]

  4. anonymous
    • 5 years ago
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    nope

  5. anonymous
    • 5 years ago
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    the its x^ 1/1-x

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  6. anonymous
    • 5 years ago
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    plz see the attachment its question number 2 -> 5th and 6th one

  7. anonymous
    • 5 years ago
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    any answers / suggestions ?

  8. nowhereman
    • 5 years ago
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    you should really learn how to use parentheses! x^1/1-x means ((x^1)/1)-x because ^ has higher priority than / (and *) which in turn has higher priority than - (and +) ! So do you mean x^(1/(1-x)) ?

  9. anonymous
    • 5 years ago
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    yup. that's right sorry for the confusion

  10. anonymous
    • 5 years ago
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    x^(1/(1-x)) looks good

  11. nowhereman
    • 5 years ago
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    So you should apply the definition of a^b

  12. anonymous
    • 5 years ago
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    ok.. first the question is clearly asking about the limits, not the derivatives ( I think it's just a typo mistake).. I will start with the first one

  13. anonymous
    • 5 years ago
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    well in l'hopital's rule u have to take drivatives

  14. anonymous
    • 5 years ago
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    Consider the function: \[y=x ^{{1 \over 1-x}}\], which is the function we are going to find its limit as it goes to 1.. we can't use l'hopital's rule since it's not in the form of (0/0) or (infinity/infinity).. so we are going to take the ln for for both side to get: \[\ln y={1 \over x-1}\ln x={\ln x \over x-1}\] (0/0 type).. now we can apply the rule find the limit for (lny) for now

  15. nowhereman
    • 5 years ago
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    Aha, applying ln, so where does that come from? Magic, magic. Just use the definition \[a^b = e^{b\ln a}\] and you will get exactly this result.

  16. anonymous
    • 5 years ago
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    hmm..

  17. anonymous
    • 5 years ago
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    sorry for the mistake.. it's 1/(1-x), so: \[\lim_{x \rightarrow 1}\ln y=\lim_{x \rightarrow 1}{\ln x \over 1-x}=\lim_{x \rightarrow 1}{1/x \over -1}=-1\] so this\[\lim_{x \rightarrow 1}x ^{{1 \over 1-x}}=e ^{\lim_{x \rightarrow 1 \ln y}}=e ^{-1}={1 \over e}\] the limit of ln y, which implies that

  18. anonymous
    • 5 years ago
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    there is a typo!! but the idea is still there.. does that make any sense to you?

  19. anonymous
    • 5 years ago
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    just because we took ln for the limit of ln y.. to get the limit of y, we just took e to the power of that first limit.. as simple as that!!

  20. anonymous
    • 5 years ago
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    i mean the equation is in 0 form right

  21. anonymous
    • 5 years ago
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    what do you mean by 0 form?

  22. anonymous
    • 5 years ago
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    why r u using ln in this

  23. anonymous
    • 5 years ago
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    It's the way to finding the limit using l'hopital's rule.. this method is well known.

  24. anonymous
    • 5 years ago
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    I am sure your instructor will use the same method to find the limit :P

  25. nowhereman
    • 5 years ago
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    The original term is not in a form where you can apply l'hospitals rule. So you write it as \[e^{\frac{\ln x}{1-x}}\] where the exponent is in the form "0/0". Then you find the limit of the exponent with l'hopsital as -1 and thus have 1/e as the end result because the exponential function is continuous: \[\lim_{x → 1} x^{\frac 1 {1-x}} = \lim_{x→1} e^{\frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\ln x}{1-x}} = e^{\lim_{x→1} \frac{\frac{1}{x}}{-1}} = e^{-1} = \frac{1}{e}\]

  26. anonymous
    • 5 years ago
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    ^^ COOL :)

  27. anonymous
    • 5 years ago
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    what about second question

  28. nowhereman
    • 5 years ago
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    What is the value that x goes to? And do you mean x^2 - ( (e^x) / x ) ?

  29. anonymous
    • 5 years ago
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    x-> infinity , and yes ur right. it was my writing mistake. as before thanks

  30. anonymous
    • 5 years ago
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    x^2 - ( (e^x) / x ) is right!

  31. nowhereman
    • 5 years ago
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    Then it diverges to -∞, because e^x grows faster than any polynomial. So for any C ϵ ℝ you can find a X so that for all x > X you have: \[e^x \geq \underbrace{x^3-Cx}_{\text{polynomial}} ⇔ x^2 - \frac{e^x}{x} \leq C\]

  32. anonymous
    • 5 years ago
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    ok

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