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Math Help - What am I doing wrong in this limit calculation?

  1. #1
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    What am I doing wrong in this limit calculation?

    Hello everyone,

    I have tried to calculate the following limit of a function, but my result doesn't match the one that my computer gives me.

    \lim \limits_{x\rightarrow-\infty}x(\sqrt{1+x^2}-\sqrt{2+x^2})

    First I multiply by \frac{\sqrt{1+x^2}+\sqrt{2+x^2}}{\sqrt{1+x^2}+\sqr  t{2+x^2}}, which gives me

    \lim \limits_{x\rightarrow-\infty}\frac{x(1+x^2-2-x^2)} {\sqrt{1+x^2}+\sqrt{2+x^2}}

    Next I simplify the numerator and factor x out of the roots in the denominator

    \lim \limits_{x\rightarrow-\infty}\frac{-x} {x(\sqrt{\frac{1}{x^2}+1}+\sqrt{\frac{2}{x^2}+1})}

    x cancels out, which gives me

    \lim \limits_{x\rightarrow-\infty}\frac{-1} {\sqrt{\frac{1}{x^2}+1}+\sqrt{\frac{2}{x^2}+1}}=-\frac{1}{2}

    My computer algebra system however tells me that the limit is 1/2. It would be -1/2 if x were approaching positive infinity, so I my problem is somehow related to x approaching negative infinity. But what exactly is it that I am doing wrong?
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  2. #2
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    For x positive, the function \sqrt{x} is defined as the unique positive real number y such that y^2=x. Hence, when x is negative, \sqrt{x^2}=-x.. The limit is taken when x tends to -\infty. Thus, in the denominator you must extract -x instead of x in each one of the square roots.
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  3. #3
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    Thank you for your help! It does work, when I factor -x out of the square root instead of x.

    However there is still one more thing that I would like to ask in reference to this problem. For a limit with x approaching a negative number, when do I have to factor -x out of a square root and when do I have to factor out x?
    The reason I ask is, because for \lim\limits_{x\rightarrow-\infty}\sqrt{1+x^2} factoring out -x would be wrong, since this function converges to positive infinity and not to negative infinity.
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  4. #4
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    You are welcome!

    Simply apply \sqrt{x^2}=|x| (this works always). In your last question, when x\to-\infty, |x|=-x and tends to +\infty as x\to -\infty
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  5. #5
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    Right, when x\rightarrow-\infty, replacing the factored out |x| with -x should work, since as \lim\limits_{x\rightarrow-\infty}-x=-(-\infty)=+\infty.
    Thanks once again.
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