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Math Help - Limits at negative infinity

  1. #1
    Junior Member ImaCowOK's Avatar
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    Limits at negative infinity

    Can someone explain, in a way that's easy for someone new to limits to understand, limits at negative infinity.

    For example:

    Find the limit of (8x - 13)/(x≥ + 7x - 17) as x approaches -∞.

    How would that be solved compared to how it would be solved if x was approaching ∞?
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  2. #2
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    Limit at negative infinity is the value that f(x) is approaching when x is decreasing without bound.

    An upper bound of the function is given by (9x)/(x^4) for x>7/17 and note that the function is positive for all x. Put these together we get
    \displaystyle 0<\frac{8x-13}{x^3+7x-17}<\frac{9x}{x^3}=\frac{9}{x^2}.
    Now if x\to -\infty, clearly 9/(x^2)\to 0. By the Squeeze Theorem, the limit at negative infinity of your function must be 0. The same inequality can be applied to the +\infty case.
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  3. #3
    Ted
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    Hello
    In both cases (infinity or -infinity), you devide the denominator and the numerator by the highest power in the denominator
    In your problem, the highest power in the denominator is x^3
    So by deviding top and bottom by x^3 we will get:

    \lim_{x\to - \infty} \dfrac{ \frac{8}{x^2} - \frac{13}{x^3} } { 1 + \frac{7}{x^2} - \frac{17}{x^3} }

    By remembering that \lim_{x\to - \infty} \dfrac{a}{x^b} = 0 where a & b are constants and b>0, you see that the final answer is 0/1 = 0.
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