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Math Help - Indeterminate Forms: Limits infinite... help

  1. #1
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    Indeterminate Forms: Limits infinite... help

    Find the limit if it exists:
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the limit if it exists:
    Now i dont know if i'm right, but maybe my answer can at least point you in the right direction.

    Lim(x--> inf+) [sin(2/x)]/(1/x)
    = Lim(x--> inf+) (2/2)[sin(2/x)]/(1/x)
    = Lim(x--> inf+) 2[sin(2/x)]/(2/x)

    Now as x--> infinity, 2/x -->0, so the above is equivalent to:
    Lim(a-->0) 2sin(a)/a
    = Lim(a-->0+) 2(sin(a)/a)
    = 2*Lim(a-->0+) sin(a)/a
    = 2*1 ..............................since lim(x-->0) sinx/x = 1
    = 2

    What bothers me is the whole approaching from the right thing, i dont know how that'll affect my solution
    Last edited by Jhevon; March 4th 2007 at 12:23 AM.
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the limit if it exists:
    Here is one way.

    Lim(x--> +infinity)[sin(2/x) / (1/x)]
    = sin(2/inf.) / (1/inf)
    = sin(0) / (0)
    = 0/0 ------------indeterminate.

    Use the L'Hopital's Rule:
    Lim(x-->a)[f(x) / g(x)] = Lim(x-->a)[f'(x) / g'(x)] for 0/0 limits.

    So,
    = Lim(x-->+inf.)[{cos(2/x) *(-2 /x^2)} / {-1 /x^2}]
    = Lim(x-->+inf.)[cos(2/x) *2]
    = 2cos(0)
    = 2*1
    = 2 ----------------answer.
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  4. #4
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    Hello, ^_^Engineer_Adam^_^!

    Find the limit if it exists:

    . . . . .sin(2/x)
    . lim . --------
    x→+∞ .(1/x)

    Let u = 1/x
    . . . . . . . . . . . . . . .sin(2u) . . . . . . . . .sin(2u)
    And we have: . lim . -------- . = . lim . 2--------- . = . 2(1) . = . 2
    . . . . . . . . . . .u→0 . . .u . . . . .2u→0 . . . 2u

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