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Math Help - Lhospitals rule helppleaseee...

  1. #1
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    Unhappy Lhospitals rule helppleaseee...

    Help with evaluating the limit of these two problems... am stuck w/ both
    Last edited by ^_^Engineer_Adam^_^; March 1st 2007 at 03:18 AM.
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Help with evaluating the limit of these two problems... am stuck w/ both
    Hello Engineer_Adam,

    1.) lim(tan(x)/ln(cos(x)), x -> Pi/2)

    If you plug in Pi/2 it's undefined;

    L'Hopital's:

    (tan(x))' = sec^2(x)

    (ln(cos(x))' = -tan(x)

    Take limit again; plug in Pi/2 --> undefined;

    L'Hopital's:

    (sec^2(x))' = (2*sin(x))/((cos(x))^3)

    (-tan(x))' = -sec^2(x)

    Repeat above;

    Undef.

    By now, it should be obvious that it will always be undefined regardless of how many times you take the derivative (the func. keeps getting messier and messier, in terms of cos in the denominator (and subsequently plugging in Pi/2 will give an undefined result).

    2.) lim(x^(1/ln(x)), x -> 0)

    Note: x^(1/ln(x)) = e; thus, the limit of the above is e.
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Help with evaluating the limit of these two problems... am stuck w/ both
    2.) lim(x->0) x^(1/ln(x))

    x = e^(ln(x)), so:

    lim(x->0) x^(1/ln(x)) = lim(x->0) e^(ln(x)/ln(x)) = e
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  4. #4
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    Ohhhh i see
    so in number 1 the limit doesnt really exist... thats were i got stuck by
    applying LHR many times...
    THe answer of the classsmates of mine is zero... theyre wrong right?
    Thanks
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  5. #5
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    Hello, ^_^Engineer_Adam^_^!

    I got a different answer for #1 . . .



    . . . . . . .tan x
    .lim . . -----------
    x→½π .ln(cos x)

    Presently, the function goes to: ∞/(-∞)

    . . . . . . . . .sec²x . . . . . . . -1 . . . . . . . . . .-2
    L'Hopital: . -------- . = . --------------- . = . -------- . = . -2·csc(2x)
    . . . . . . . . -tan x . . . . .sin(x)·cos(x) . . . .sin(2x)


    . lim . [-2·csc(2x)] . = . -2(1) . = . -2
    x→½π

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, ^_^Engineer_Adam^_^!

    I got a different answer for #1 . . .



    Presently, the function goes to: ∞/(-∞)

    . . . . . . . . .sec²x . . . . . . . -1 . . . . . . . . . .-2
    L'Hopital: . -------- . = . --------------- . = . -------- . = . -2·csc(2x)
    . . . . . . . . -tan x . . . . .sin(x)·cos(x) . . . .sin(2x)


    . lim . [-2·csc(2x)] . = . -2(1) . = . -2
    x→½π

    Huh? I don't follow how you got:

    -1/[sin(x)*cos(x)] ?? You have to differentiate more since it's an indeterminate form; eventually you will see no matter how many times you differentiate it will be undefined- see explanation above.
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  7. #7
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Ohhhh i see
    so in number 1 the limit doesnt really exist... thats were i got stuck by
    applying LHR many times...
    THe answer of the classsmates of mine is zero... theyre wrong right?
    Thanks
    Yes, they are .
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AfterShock View Post
    Huh? I don't follow how you got:

    -1/[sin(x)*cos(x)] ?? You have to differentiate more since it's an indeterminate form; eventually you will see no matter how many times you differentiate it will be undefined- see explanation above.
    What he did was to simplify by making sec^2(x) = 1/cosx and tanx = sinx/cosx.

    I dont really see anything wrong with doing that, but hey, that's just me, what do i know about math?
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  9. #9
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    (tan x)/(ln cos x)=(sin x)/(cos x ln cos x)

    Now use limit laws.

    lim (x --> pi/2 - ) sin x = 1

    While,

    lim (x --> pi/2 -) (cos x ln cos x) = 0

    Thus,
    lim (sin x)/(cos x ln cos x) = BIG
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    What he did was to simplify by making sec^2(x) = 1/cosx and tanx = sinx/cosx.

    I dont really see anything wrong with doing that, but hey, that's just me, what do i know about math?
    Well 1/(cos^2(x)); yes, I realize the simplification. But, that does not equal -1/(sin(x)*cos(x)).
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AfterShock View Post
    Well 1/(cos^2(x)); yes, I realize the simplification. But, that does not equal -1/(sin(x)*cos(x)).
    sec^2x/-tanx = (1/cos^2x)/(-sinx/cosx) = (1/cos^2x)*(-cosx/sinx) = -1/sinxcosx
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  12. #12
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    I like to mention something about,

    tan (x)/ ln( cos x) near point x=pi/2

    Note that if x>pi/2 and small enough.
    Then cos x<0 and we have a problem.

    Thus, the limit at x=pi/2
    tan (x)/ln (cos x)

    Because there does not exists an open interval for which the function is defined (except possibly at x=pi/2).

    Therefore, we can only consider the left-handed limit.
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  13. #13
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    Ah, yes. My fault, I was dealing with derivatives and not paying attention to what you were trying to show. Yes.

    Now, what's the limit of -1/(sin(x)*cos(x)) ... undefined. The -2 is what through me off.
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  14. #14
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    Quote Originally Posted by ThePerfectHacker View Post
    I like to mention something about,

    tan (x)/ ln( cos x) near point x=pi/2

    Note that if x>pi/2 and small enough.
    Then cos x<0 and we have a problem.

    Thus, the limit at x=pi/2
    tan (x)/ln (cos x)

    Because there does not exists an open interval for which the function is defined (except possibly at x=pi/2).

    Therefore, we can only consider the left-handed limit.
    Precisely, since the left hand limit does not equal the right hand limit, the lim does not exist.
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  15. #15
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    Quote Originally Posted by AfterShock View Post
    Precisely, since the left hand limit does not equal the right hand limit, the lim does not exist.
    However, L'Hopital limit still applies for one-sided differenciable function in the neigborhood of the point.
    In that case the limit still does not exist.

    Just be show a picture.
    Look here.
    Attached Thumbnails Attached Thumbnails Lhospitals rule helppleaseee...-picture8.gif  
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