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Math Help - limits at negative infinity

  1. #1
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    limits at negative infinity

    limits at negative infinity-picture-19.png

    when taking the limit as x goes to negative infinity, and we divided by negative sqrt(x^2) why does the term 5x not become -5?

    the negative is associated only with the second term in the denominator but not the 5, why?

    also why is the denominator not divided by a negative?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: limits at negative infinity

    This is what happens:
    \lim_{x \to \infty} \frac{5x+\sqrt{x^2+2x}}{4+x} = \lim_{x \to \infty} \frac{5x+\sqrt{x^2\left(1+\frac{2}{x}\right)}}{x  \left(\frac{4}{x}+1\right)}
    = \lim_{x \to \infty} \frac{5x+|x|\sqrt{1+\frac{2}{x}}}{x\left(\frac{4}{  x}+1\right)}

    Suppose x \to +\infty then |x| = x thus
    \lim_{x \to +\infty} \frac{x\left(5+\sqrt{1+\frac{2}{x}}\right)}{x\left  (\frac{4}{x}+1\right)} = \lim_{x \to +\infty} \frac{5+\sqrt{1+\frac{2}{x}}}{\frac{4}{x}+1} = 6

    Suppose x \to -\infty then |x|=-x
    Now (check it) the limit will be equal to 4
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: limits at negative infinity

    Perphas this also can help you: substituting for example x=-3 in \frac{5x+\sqrt{x^2+2x}}{4+x} we get \frac{5(-3)+\sqrt{(-3)^2+2(-3)}}{4+(-3)}. Now, divide numerator by -\sqrt{(-3)^2}\;(=-3) and denonimator by -3 . What do you obtain? (substtitute now -3 by x)
    Thanks from kingsolomonsgrave
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