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Math Help - Limit Problem with Infinity

  1. #1
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    Limit Problem with Infinity

    \dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}

    Now divide everything by the highest index in the denominator

    \dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}

    \dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4} - See the x^{-1} in the bottom?

    \dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4} - This becomes x^{1} in the top. But how to evaluate to get -\dfrac{1}{4}
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    Re: Limit Problem with Infinity

    Quote Originally Posted by Jason76 View Post
    \dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}
    \dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1 - 4x^{-2}}{2x^{-2} + x^{-1} - 4}
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  3. #3
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    Re: Limit Problem with Infinity

    x to the power -1 is the same as 1/x, and as x tends to become infinitely large number, 1/x tends to become 0, hence it can simply be ignored in the denominator. Actually, that jump x to the power -1 made from denominator to numerator is a complete mathematical nonsense. Denominator should be factored and only then parts of such factored denominator could 'climb' over the fraction line (with its exponent changing sign) and multiply the numerator.
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    Re: Limit Problem with Infinity

    Quote Originally Posted by Jason76 View Post
    \dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}

    Now divide everything by the highest index in the denominator

    \dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}

    \dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4} - See the x^{-1} in the bottom?

    \dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4} - This becomes x^{1} in the top. But how to evaluate to get -\dfrac{1}{4}
    No, the " x^{-1}" does NOT become "x" in the top! You cannot move just one term from denominator to the numerator like that.

    \frac{1}{{1/5}- 4} is NOT equal to \frac{5}{-4}.
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    Re: Limit Problem with Infinity

    Quote Originally Posted by Jason76 View Post
    \dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4} - See the x^{-1} in the bottom?

    \dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4} - This becomes x^{1} in the top. But how to evaluate to get -\dfrac{1}{4}
    What is your proof or what rule did you use to have this \dfrac{a}{b^{-x}+c} = \dfrac{a+b^{x}}{c} as valid?

    Also we don't write 0 + -4. You can't have 2 symbols for operations together. The correct way is 0+(-4).
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    Re: Limit Problem with Infinity

    Quote Originally Posted by Jason76 View Post
    \dfrac{\lim}{x \to \infty}\dfrac{x^{2} - 4}{2 + x - 4x^{2}}

    Now divide everything by the highest index in the denominator

    \dfrac{\lim}{x \to \infty}\dfrac{x^{2}/ x^{2} - 4/x^{2}}{2/ x^{2}+ x/x^{2} - 4x^{2}/x^{2}}

    \dfrac{\lim}{x \to \infty}\dfrac{1 -0}{0 + x^{-1} - 4} - See the x^{-1} in the bottom?

    \dfrac{\lim}{x \to \infty}\dfrac{1 - 0 + x^{1}}{0 + - 4} - This becomes x^{1} in the top. But how to evaluate to get -\dfrac{1}{4}
    Limit Problem with Infinity-12-aug-13.png
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