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Math Help - Confused about process for finding limits at infinity?

  1. #1
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    Confused about process for finding limits at infinity?

    Hi, I'm practicing for Calc test and the solutions to a problem are throwing me off. I know that to find a limit as x approaches infinity you can divide everything by the highest power x in the denominator but the solution steps show the person doing this:



    I see that they divided everything by x^2, including everything under the square root. But wouldn't you need to divide the stuff under the square by x^4 since that's sqrt(x^4) is equivalent to x^2? Any help is appreciated

    EDIT: Also the original question actually asks for the limit as x approaches NEGATIVE infinity. I'm looking at this problem after posting and I'm frankly lost on the steps, but I will keep working on it.
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  2. #2
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    I really don't understand what you are asking from us.
    The algebra in the solution is straightforward.
    So what can we say?
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  3. #3
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    \displaystyle \lim_{x\to\infty}\frac{x^2*(2x)^3}{x^4*\sqrt{x^2}}  \Rightarrow\lim_{x\to\infty}\frac{8x^5}{x^5}=8

    Just take the highest powers from each product in both the denominator and numerator and then simplify.
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  4. #4
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    edit:

    I had an error in my algebra...
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  5. #5
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    Quote Originally Posted by adkinsjr View Post
    The algebra doesn't seem simple to me. I would just use dwsmith's method of highest powers. But I'm not following the algebra in the solution either.
    Well that is why it is called Pre-Calculus.
    You need basic algebra to do calculus.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Well that is why it is called Pre-Calculus.
    You need basic algebra to do calculus.
    I've been through calcs 1-3 and I obtained A's in all 3.
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  7. #7
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    Ok, I see what your problem is now...

    Look at it like this:

    \lim_{x->\infty}\frac{x^2(2x+1)^3}{(x^2+2)^2\sqrt{x^2+2x+3  }}=\lim_{x->\infty}\frac{x^2x^3\frac{1}{x^3}(2x+1)^3}{x^5\fra  c{1}{x^4}(x+2)^2\frac{1}{x}\sqrt{x^2+2x+3}}

    This seems to be what they did, they didn't divide the square root by x^2, they just divided it by x... However, I still think they messed up after that. The algebra they did just doesn't make sense...not to me anyways...
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  8. #8
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    Quote Originally Posted by DannyMath View Post
    Hi, I'm practicing for Calc test and the solutions to a problem are throwing me off. I know that to find a limit as x approaches infinity you can divide everything by the highest power x in the denominator but the solution steps show the person doing this:



    I see that they divided everything by x^2, including everything under the square root. But wouldn't you need to divide the stuff under the square by x^4 since that's sqrt(x^4) is equivalent to x^2? Any help is appreciated

    EDIT: Also the original question actually asks for the limit as x approaches NEGATIVE infinity. I'm looking at this problem after posting and I'm frankly lost on the steps, but I will keep working on it.
    \displaystyle\frac{x^2(2x+1)^3}{(x^2+2)^2\sqrt{x^2  +2x+3}}=\frac{x^2\left[x\left(2+\frac{1}{x}\right)\right]^3}{\left[x^2\left(1+\frac{2}{x^2}\right)\right]^2\left[\sqrt{x^2}\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}\right]}

    =\displaystyle\frac{x^2x^3\left(2+\frac{1}{x}\righ  t)^3}{x^4x\left(1+\frac{2}{x^2}\right)^2\sqrt{1+\f  rac{2}{x}+\frac{3}{x^2}}}

    so there is a little error in the text but it is inconsequential to the calculation of the limit.
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\frac{x^2(2x+1)^3}{(x^2+2)^2\sqrt{x^2  +2x+3}}=\frac{x^2\left[x\left(2+\frac{1}{x}\right)\right]^3}{\left[x^2\left(1+\frac{2}{x^2}\right)\right]^2\left[\sqrt{x^2}\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}\right]}

    =\displaystyle\frac{x^2x^3\left(2+\frac{1}{x}\righ  t)^3}{x^4x\left(1+\frac{2}{x^2}\right)^2\sqrt{1+\f  rac{2}{x}+\frac{3}{x^2}}}

    so there is a little error in the text but it is inconsequential to the calculation of the limit.
    There you go, thankyou.. I see where they messed up now. I couldn't spot that...
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  10. #10
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    Quote Originally Posted by Plato View Post
    Well that is why it is called Pre-Calculus.
    You need basic algebra to do calculus.
    I don't understand what the point is in posting this type of stuff. I doubt that you believe every person who has passed Pre-Cal would find this straightforward.

    Quote Originally Posted by Archie Meade View Post
    so there is a little error in the text but it is inconsequential to the calculation of the limit.
    Thanks for the helpful info.
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  11. #11
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    Hi again all. I'm doing this problem again but this time it's as x goes to negative infinity and I see that the graph shows that as x goes to negative infinity, the limit is -8. Is this because when you are going from sqrt(x^2) to x you need to add a negative sign to keep the expression true? Because that's the only way I found I would get -8 rather than 8. Thanks again
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  12. #12
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    Look carefully at the original problem.
    There are four factors: x^2,~(2x+3)^3,~(x^2+1)^2~\&~\sqrt{x^2+2x+3}.
    As x\to -\infty all but one of those factors are positive, one is negative.
    So the answer is negative. The algebra part of the workings simply helps us find the numerical value. See post #3. But it does not give the correct sign. That comes from an analysis of the original factors.
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  13. #13
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    \sqrt{x^2} = -x for negative values of x.

    In general, \sqrt{x^2} =|x|
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  14. #14
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    Quote Originally Posted by DrSteve View Post
    \sqrt{x^2} = -x for negative values of x.

    In general, \sqrt{x^2} =|x|
    Yes, the equation contains the positive square root.
    See also post 12 for the alternative evaluation of the sign.

    Post 8 takes x as positive only.
    Last edited by Archie Meade; December 13th 2010 at 12:44 PM. Reason: misread a post
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  15. #15
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    Yes - that's true. The equation does contain the positive square root. What I've given is the definition of the positive square root of x (note that -x is a positive number when x<0).

    If you're using the shortcut method to compute the limit, then this information is not needed. But if you want to write out the details more rigorously (as was done in the very first post), then this definition will be needed.
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