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seancashmere

  • 3 years ago

More help with l'hospital's rule: can I evaluate a problem using l'hospital's rule when the following condition exists: lim(x-->infinity) (xe^(1/x)-x)

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  1. seancashmere
    • 3 years ago
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    Sorry, the condition, after I worked the problem is -1/infinity

  2. hartnn
    • 3 years ago
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    no, L'Hopital's can be applied only if you have the form 0/0 or \(\infty\)/\(\infty\) so, what you can do is, (xe^(1/x)-x) =x( (e^(1/x)-1)) = (e^(1/x)-1) / (1/x) now its 0/0 form and you can apply L'Hopitals.

  3. kirbykirby
    • 3 years ago
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    you get a form infinity - infinity which is undefined

  4. seancashmere
    • 3 years ago
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    Thank you, I set it up differently :(

  5. seancashmere
    • 3 years ago
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    Also, aren't there other conditions for l'hospital's? For example 1^infinity?

  6. hartnn
    • 3 years ago
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    if you have 1^infinity form, you can apply log to both sides to get 0/0 form, but there are other ways to deal with 1^infinity form also.

  7. hartnn
    • 3 years ago
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    for L'Hopital's , only 0/0 or infinity/infinity should be there.

  8. kirbykirby
    • 3 years ago
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    You can only apply L'Hopital's rule when you have the form \[0/0\] or \[\infty/\infty\]. When you have 1^infinity, you usually transform your by using ln (natural log) on both sides of your equation (set your limit as y = lim... blabla) then ln both sides

  9. seancashmere
    • 3 years ago
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    I got infinity over infinity when I worked that problem, how did you get zero over zero?

  10. hartnn
    • 3 years ago
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    i meant 0/0 or infinity / infinity ...any one of the form you can get...

  11. seancashmere
    • 3 years ago
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    I know, but after evaluating the above problem, I got -infinity/infinity, you said you got 0/0

  12. seancashmere
    • 3 years ago
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    How did you work it differently?

  13. hartnn
    • 3 years ago
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    oh, 1/x =0, e^0 =1 1-1=0 <--numerator 1/x=0 <--denom.

  14. seancashmere
    • 3 years ago
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    You are right! Thank you!!

  15. seancashmere
    • 3 years ago
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    Thanks for the help. Keep your eyes peeled, I may need help differentiating it. Soon. I'll close this for now and open a new one if I need help differentiating it.

  16. kirbykirby
    • 3 years ago
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    (e^(1/x)-1) / (1/x)\[\lim_{x \rightarrow \infty} \frac{e^{1/x}-1}{\frac{1}{x}}= \frac{\lim_{x \rightarrow \infty}e^{1/x}-1}{\lim_{x \rightarrow \infty}\frac{1}{x}}=\frac{e^{1/\infty}-1}{\frac{1}{\infty}}=\frac{e^0-1}{0}=0/0\]

  17. hartnn
    • 3 years ago
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    ok, sure :)

  18. seancashmere
    • 3 years ago
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    Thanks Kirby!

  19. hartnn
    • 3 years ago
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    one tip, before differentiating, substitute x =1/y so, that y-->0 then diff. num and denom. seperately. it'll much easier than to diff. by x.

  20. seancashmere
    • 3 years ago
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    What? Why? I don't understand your tip at all... does infinity equal 1/0?

  21. kirbykirby
    • 3 years ago
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    It's the limit as x->0 of 1/x ... it's different than just "1/x" where x=0

  22. hartnn
    • 3 years ago
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    yes, it does.

  23. kirbykirby
    • 3 years ago
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    |dw:1358746300788:dw|

  24. hartnn
    • 3 years ago
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    x---> infinity x is very very large but not equal to infinity 1/x will be very very small, but not =0 -----> 1/x->0

  25. seancashmere
    • 3 years ago
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    interesting

  26. kirbykirby
    • 3 years ago
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    As you go towards x->0 on the x-axis, you approach infinity because there is an asymptote at x=0. So, the graph is continuously stretching out infinitely along the asymptote

  27. seancashmere
    • 3 years ago
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    It's hard for me to visualize. Why is that fraction easier to use than just x as it approaches infinity?

  28. seancashmere
    • 3 years ago
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    So I have (-e^(1/x)/x^2)/-(1/x^2), which, using l'hospital's rule gives me 0/0. I differentiate that and I have

  29. kirbykirby
    • 3 years ago
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    x->0 from the right obviously, it approaches +infinity, but -infinity if it approaches 0 from the left

  30. seancashmere
    • 3 years ago
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    1/infinity... which is zero, what do you think? Am I differentiating it correctly?

  31. hartnn
    • 3 years ago
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    see, if you differentiate directly, you'll need to differentiate, 1/x in the denominator, which will give you -1/x^2 , which again will be 0 for x->infinity. BUT, if you put y=1/x you'll have y in the denominator, and differentiating y will give u 1. got the point ?

  32. seancashmere
    • 3 years ago
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    no... give me a second for my brain to process this

  33. seancashmere
    • 3 years ago
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    Yeah, I don't see how putting zero in a denominator will ever give me one?

  34. seancashmere
    • 3 years ago
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    if y is approaching zero, that is, and is in the denominator

  35. hartnn
    • 3 years ago
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    'putting zero in a denominator ' ??

  36. seancashmere
    • 3 years ago
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    you said I should use y--->0 instead of 1/x as x---> infinity. If I plug y in my denominator, then I'll have zero.

  37. seancashmere
    • 3 years ago
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    I'm supposing I actually have Zero idea what you're talking about... in the numerator

  38. hartnn
    • 3 years ago
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    wait, did you try differentiating without any substitution ? then you'll see what problems arise. then you put 1/x = y.

  39. seancashmere
    • 3 years ago
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    I tried differentiating and I wound up with 1/infinity, which is zero. I'm wondering if I got it or if I botched it.

  40. hartnn
    • 3 years ago
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    umm....no, you don't get 1/infinity....show your steps....

  41. seancashmere
    • 3 years ago
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    okay. I have to differentiate (-e^(1/x)/X^2)/(1/(x^2))

  42. seancashmere
    • 3 years ago
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    I get -e^(1/x)/(x^2) * -x^2

  43. hartnn
    • 3 years ago
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    wait.

  44. seancashmere
    • 3 years ago
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    that gives me -e^(1/x)/x^2 * -1

  45. seancashmere
    • 3 years ago
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    I'm waiting

  46. kirbykirby
    • 3 years ago
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    (-e^(1/x)/X^2)/(1/(x^2))= -e^(1/x)

  47. hartnn
    • 3 years ago
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    numerator = e^(1/x) -1 ---> (-1/x^2) e^(1/x) denominator =1/x ------> (-1/x^2) overall = e^(1/x) yeah...

  48. seancashmere
    • 3 years ago
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    so I do that before I differentiate

  49. hartnn
    • 3 years ago
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    you got (-1/x^2) e^(1/x) and (-1/x^2) after differentiating only...

  50. hartnn
    • 3 years ago
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    and then you can directly put 1/x = 0, e^0=1

  51. seancashmere
    • 3 years ago
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    so I have 1/0, or indeterminate?

  52. hartnn
    • 3 years ago
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    you have 1 in the denominator now...not 0

  53. seancashmere
    • 3 years ago
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    got it, so my answer is 1!

  54. hartnn
    • 3 years ago
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    yes!

  55. seancashmere
    • 3 years ago
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    e^0/1?

  56. seancashmere
    • 3 years ago
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    vunderbar! so I shouldn't have differentiated in the first place.

  57. seancashmere
    • 3 years ago
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    I just evaluated it wrong. You are amazing. Who are you? Thanks a million!

  58. hartnn
    • 3 years ago
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    i am just another helper, glad i could help :) welcome ^_^

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