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Math Help - Limits at infinity?

  1. #1
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    Limits at infinity?

    Can anyone help me with these limits

    1. As x approaches infinity

    ((sin2x))/((x))

    2. As x approaches infinity

    ((1))/((2x+sin(x))

    3. As x approaches infinity

    ((x+ square root(x^2+3)) I have some work for this one

    ((x+ square root(x^2+3))/((1)) I rationalized the function

    ((x^2+x^2+3))/((x+square root(x^2+3))

    Now my main issue is how would I get a common denominator what should I chose?
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  2. #2
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    Re: Limits at infinity?

    For the first one:
    \lim_{x\to \infty} \frac{\sin(2x)}{x}

    If x \to \infty , then \sin(2x) goes to? ... (think of the range)
    Last edited by Ackbeet; August 1st 2011 at 06:29 AM.
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  3. #3
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    Re: Limits at infinity?

    It seems for my first question the limit would be zero.
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    Re: Limits at infinity?

    Quote Originally Posted by homeylova223 View Post
    2. As x approaches infinity
    ((1))/((2x+sin(x))
    If x\ge 3 then 0<2x-1\le 2x+\sin(x)\le 2x+1.

    That gives \frac{1}{2x+1}\le \frac{1}{2x+\sin(x)} \le \frac{1}{2x-1}

    If x\to \infty what happens to  \frac{1}{2x+\sin(x)} ~?
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Limits at infinity?

    Quote Originally Posted by homeylova223 View Post
    It seems for my first question the limit would be zero.
    Correct!
    For the last one, make a difference by calculating x \to +\infty and  x \to -\infty .
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  6. #6
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    Re: Limits at infinity?

    Seeing Plato inequality the limit would have to approach x at zero.

    Now all I have left is my third question.
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  7. #7
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    Re: Limits at infinity?

    Can you go further with the hint I gave?
    \lim_{x\to + \infty} \left(x+\sqrt{x^2+3x}\right)=+\infty+\infty=\infty
    For x \to - \infty you'll get the indetermined form \infty-\infty so multiply and divide the limit by x-\sqrt{x^2+3} ...
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  8. #8
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    Re: Limits at infinity?

    Hmm I am slight confused

    if I multiply by ((x+square root(x^2+3))

    I get ((x^2+x^2+3))/((x+square root(x^2+3)) I used the denominator x^2 for the numerator

    ((x^2+x^2+3))/((x^2)) to get ((1+1+3/x^2))

    But what could I use for the denominator I tried x but I think it incorrect.
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  9. #9
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    Re: Limits at infinity?

    What do you think of:
    \lim_{x\to -\infty} \frac{(x+\sqrt{x^2+3})\cdot (x-\sqrt{x^2+3})}{x-\sqrt{x^2+3}}=\lim_{x\to -\infty}\frac{x^2-x^2-3}{x-\sqrt{x^2+3}}=\lim_{x\to -\infty}\frac{-3}{-\infty-\infty}=\lim_{x\to -\infty} \frac{-3}{-(\infty+\infty)}=0

    Do you understand?
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  10. #10
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    Re: Limits at infinity?

    I am wondering when you get ((x^2-x^2-3))/((x-square root(x^2+3))

    How do you get -infinity-infinity
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  11. #11
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    Re: Limits at infinity?

    Quote Originally Posted by homeylova223 View Post
    I am wondering when you get ((x^2-x^2-3))/((x-square root(x^2+3))

    How do you get -infinity-infinity
    You don't, that is an invalid move.

    You have:

    \lim_{x \to \infty}\frac{-3}{x-\sqrt{x^2+3}}=\lim_{x \to \infty}\frac{-3}{x-x\sqrt{1+3/x^2}}

    So now you need only worry about:

    \lim_{x \to \infty}\left(x-x\sqrt{1+3/x^2}\right)

    and this can be determined using a power series expansion of the square root.

    CB
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