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Simple limit questions help (work is there)

  1. Jan 28, 2005 #1

    mad

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    Hello, I have those limit question to answer. (I applied Hospital Law)


    1) lim cotgx / lnx
    x-> 0+

    Heres what I did:

    1) form infinite/infinite;

    -(cosec x)^2 / (1/x) = -2cosecx*-cosecx*cotgx / (-1x^-2) = -2(cosecx)^2 *cotgx * x^2 = 0
    (wrong answer, I dont understand what's wrong with what I did)


    2nd question: I got the answer right but I want to know if I can do what I did or it was just luck..

    lim x^100 / e^x
    x-> +oo (inf.)

    what I did: y = e^x, ln y = x. so
    x^100 / y = ln x^100 / lny = 100lnx/lny = 100/x = 0


    and question 3)

    lim ln(sinx) / ln(tgx)
    x-> 0+

    what I did:
    (cosx/sinx) / (sec x)^2 = cosx/sinx * 1/(sec x)^2 = (cosx)^3 / sinx
    = -3 cosx *sinx (derivat.) = -3*0 = 0 (wrong answer, answer is 1 from the booK)


    Any help would be greatly appreciated
     
  2. jcsd
  3. Jan 29, 2005 #2

    mad

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    Please someone.. they're simple limits..
     
  4. Jan 29, 2005 #3

    Curious3141

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    First differentiation is correct. After that, reexpress like so and apply LH again:

    [tex]-\frac{\csc^2{x}}{\frac{1}{x}} = -\frac{x}{\sin^2{x}}[/tex]

    I'd have to say luck, because honestly, I don't know what you're doing there.

    Try this : you need to apply LH 100 times to get the numerator expression to a constant. What happens when you differentiate [itex]x^n[/tex] n times ? What happens when you differentiate the denominator a 100 times ?


    Error in differentiating the denominator : You forgot that it was [itex]\ln(\tan x)[/itex] and not just [itex]\tan x[/itex]
     
  5. Jan 29, 2005 #4

    mad

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    Thanks a lot for the help :).

    For the 2nd problem, I'll explain more in details so maybe what I did is okay..

    lim x^100 / e^x
    x-> +oo (inf.)

    what I did:
    I posed y = e^x , to simplify the writing. If y = e^x , then ln y = x

    So I replace that in the equation:

    lim x^100 / y (since y= e^x)
    x-> +oo

    Then I can do this:
    ln x^100 / ln y

    I simplified:

    100 lnx/ln y

    Then I apply LH's law:

    = (100/ x)/ 1 (since lny = x and x' = 1)
    So that makes the limit 100 / infinite = 0

    Is what I did okay?
     
  6. Jan 29, 2005 #5

    dextercioby

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    No.Absolutely wrong...As Courious said,since it's a limit +infty/+infty (assumming "n" to be positive & real),by applying "n" times the derivative (from l'Hôspital's rule),u get the limit of the ratio between the factorial of "n" (in the case of "n" natural) and the exp of "x".

    Daniel.
     
  7. Jan 29, 2005 #6

    mad

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    I really can't understand what you're saying, not because it is unclear, but because english isn't my first language.
    So you are saying to apply the limit 100 times? That would be a big number / infinity which would be 0 . Is that what you are saying? What if the denominator wasnt e^x, which when I derivate it always gives e^x...
    thanks a lot for the help
     
  8. Jan 29, 2005 #7

    learningphysics

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    Can you explain how you did this step? Why did you take ln of the numerator and denominator?
     
  9. Jan 29, 2005 #8

    dextercioby

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    The English is not my native language,too. :smile:

    Yes,"n" times the derivative on the numerator will give u a constant.Since the exponential derived "n" times is still the exponential (which diverges in that limit),then the ratio goes to zero...

    As for the second part,it would be different,if the function would not be "exponential type"...

    Daniel.
     
  10. Jan 29, 2005 #9

    mad

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    Well, I thought that since I can multiply both the numerator and denominator by something, that I could apply ln to both, like in an equation, but I was wrong..

    I got that from my teacher's notes. He was doing a limit of x^x , and used ln to find it, it was a long demonstration, but he didnt explain it yet because the class was finished, so I tried it by myself. I'll know what I can do when he explains it :)
     
  11. Jan 29, 2005 #10

    dextercioby

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    It's good thing that u eventually realized that in general:
    [tex] \ln(\frac{a}{b})\neq \frac{\ln a}{\ln b}[/tex]

    Daniel.
     
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