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Indeterminate Form_Limits

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the following limit:

    lim x-> 0+ ((1/x)^(tan x))


    2. Relevant equations



    3. The attempt at a solution
    This gives me the indeterminate form infinity raised to power zero.
    After trying two methods I still end up with an indeterminate form.

    e^ ln (1/x)^(tan x)= e^[ln (1/x) * (tan x)]
    (e^u) * d/dx (ln (1/x) (tan x))
    (1/x)^(tan x) * d/dx (ln (1/x) (tan x))
    (1/x)^(tan x) * [(ln(1/x)sec^2(x)) + (1/x)(tan x))]
    That gives me an indeterminate form.

    My second method:
    ln f(x)= (tan x) * (ln (1/x))
    f(x)= e^[(tan x) * (ln (1/x))]
    lim x->0+ (1/x)^(tan x)= e ^ [lim x->0+ ((tan x) * (ln (1/x)))]
    Also, what I think is an indeterminate form.
    I'm inclined to say the answer is one, but I know that infinity to the power zero is invalid, and indeterminate. I'm stumped. Help anyone? Thanks so much!
     
  2. jcsd
  3. Aug 4, 2013 #2

    lurflurf

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    Be careful it is possible to make matters worse. I am not able to follow your work completely. Taking the first attempt we have
    $$\lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{\tan(x)}= \lim_{x \rightarrow 0^+} \exp \, \log \left( \frac{1}{x} \right)^{\tan(x)}\\
    \text{The exp can be moved out of the limit as it is continuous. We want $\infty/\infty$ form like this} \\
    = \exp \lim_{x \rightarrow 0^+} -\log(x)/\cot(x)\\
    $$
    Apply l'Hôpital's rule. At each application of l'Hôpital's rule look to rearange. Multiple applications of l'Hôpital's without care can complicate matters. See if you can finish.
     
  4. Aug 4, 2013 #3
    Yes sir, that is what I have. Using the natural log to get rid of the indeterminate form of infinity to the power zero. Sorry it was unclear.

    I tried using L'hopital's rule but I still get indeterminate forms of infinity/infinity.

    (-1/x)/(-csc^2(x))
    (1/(x^2)/(2 cot x* csc^2(x)) (L'Hopital's Rule)
     
  5. Aug 4, 2013 #4

    lurflurf

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    Yes as I hinted you want to rearrange the equation at this point
    $$(1/x)/\csc^2(x)=\sin^2(x)/x=x(\sin(x)/x)^2=(1/2)(1-\cos(2x))/x$$
    Any of these rearrangements will work much better. Do you know the limit of sin(x)/x?
     
  6. Aug 4, 2013 #5
    Yes I believe that it is zero.
    And I understand now how you used those trigonometric identities in the last steps.
    So since I get a 0/0 indeterminate form in the last step again, I use L'Hopital's if I am not mistaken, and I have:

    sin(2x)/1= sin(2x). Then when lim x->0+ I have sin(2x)= sin(0)= 0?

    Thank you.
     
  7. Aug 4, 2013 #6
    I'm not sure if what I have here is correct. Can anyone please confirm?
     
  8. Aug 4, 2013 #7

    lurflurf

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    Yes that is right. Remember the original integral was e^ new one so L=e^lim sin(2x)
     
  9. Aug 4, 2013 #8
    Yes, so e^0= 1. Thank you for all your help. I really appreciate it.
     
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