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Math Help - new to calculus--test monday--limits

  1. #1
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    new to calculus--test monday--limits

    the limit as x approaches infinity of the square root of (4x^2 + 1)/8x + 7 equals....

    This is what I have before I get lost...

    1) I multiply by the conjugate to get 4x^2 + 1/8x+7 square root (4x^2 + 1)

    2)then I multiply the numerator and denom by (1/x) to get (4x + 1/x) / 8 + 7/x square root (1/x^2 times (4x^2 + 1)

    3) then I know that as x approaches infinity 1/x nears 0, 7/x nears 0, and square root of 1/x^2 nears 0, so I'm left wtih 4x/8? I have no clue!! help? what do I do when the numerator has a higher degree than the denom? any helpful hints in thinking this through?
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  2. #2
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    Quote Originally Posted by obsmith08 View Post
    the limit as x approaches infinity of the square root of (4x^2 + 1)/8x + 7 equals....

    This is what I have before I get lost...

    1) I multiply by the conjugate to get 4x^2 + 1/8x+7 square root (4x^2 + 1)

    2)then I multiply the numerator and denom by (1/x) to get (4x + 1/x) / 8 + 7/x square root (1/x^2 times (4x^2 + 1)

    3) then I know that as x approaches infinity 1/x nears 0, 7/x nears 0, and square root of 1/x^2 nears 0, so I'm left wtih 4x/8? I have no clue!! help? what do I do when the numerator has a higher degree than the denom? any helpful hints in thinking this through?

    when dealing with limits going to infinity, if the numerator has a higher degree then the denom, then the limit is going to go to infinity. I think you made this problem a little bit harder then it needs to be, the way i would look at it is ignore all the constants in this problem and look at sqrt(4x^2)/8x, then u cancel out the sqrt and the number inside squared getting u, 2x/8x then no matter what the x value goes to in the limit u are going to be left with 2/8, so i would say the answer is 2/8 for that problem. Im a freshman in college so i might be wrong but i would say im 75% confident with what im telling u.
    Last edited by ahawk1; January 31st 2009 at 10:55 AM.
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  3. #3
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    Is this it:

    \lim_{x\to {\infty}}\frac{\sqrt{4x^{2}+1}}{8x+7}

    If so, divide the top by \sqrt{x^{2}} and the bottom by x.

    We get:

    \frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{1}{x^{2}}}}  {\frac{8x}{x}+\frac{7}{x}}

    Now, we can see we end up with \frac{2}{8}=\frac{1}{4} as the limit.
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