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Math Help - Stuck on a limit problem

  1. #1
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    Stuck on a limit problem

    Hey everyone, I'm looking for help on how to solve this limit problem.

    This is the information provided, sorry about the way I wrote it out.
    f(x)=(x^3)-(x^2)-(3x)-1

    h(x)=(f(x))/(g(x))

    h is continuous except at x=-1
    lim x ->infinity of h(x)= infinity
    lim x-> -1 of H(x)= 1/2

    Find the value of lim x->0 of (3h(x)+f(x)-2g(x))

    I've tried factoring out f(x) and using it in lim x->-1 of h(x)=(f(x))/(g(x)), but I keep arriving with 0 being the numerator, which can't be right since the limit's supposed to be 1/2. Any and all help would be appreciated, thanks.
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  2. #2
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    Quote Originally Posted by crashen View Post
    Hey everyone, I'm looking for help on how to solve this limit problem.

    This is the information provided, sorry about the way I wrote it out.
    f(x)=(x^3)-(x^2)-(3x)-1

    h(x)=(f(x))/(g(x))

    h is continuous except at x=-1
    lim x ->infinity of h(x)= infinity
    lim x-> -1 of H(x)= 1/2

    Find the value of lim x->0 of (3h(x)+f(x)-2g(x))

    I've tried factoring out f(x) and using it in lim x->-1 of h(x)=(f(x))/(g(x)), but I keep arriving with 0 being the numerator, which can't be right since the limit's supposed to be 1/2. Any and all help would be appreciated, thanks.
    Since f(x)=x^3-x^2-3x-1=(x+1)(x^2-2x-1)\,\,but\,\,h(x) is not continuous at -1, we must have that g(-1)=0 , and as h(x) \xrightarrow [x\to \infty] {} \infty\,\,but\,\,h(x) \xrightarrow [x\to -1] {} \frac{1}{2}, if g(x) is a polynomial function then g(x)=(x+1)k(x), with k being a polynomial function of degree \leq 1\,\;s.t.\,\;k(-1)=4 , and as h is continuous everywhere but -1 it must be that k has no real roots, so k(x)=4, and we can say that:

    h(x)=\frac{x^2-2x-1}{k(x)}=\frac{x^2-2x-1}{4}\,,\;for\,\,x\neq -1

    Now substitute stuff and evaluate your limit.

    Tonio
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