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Math Help - Abstraction and Indeterminate Forms of Limits

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    Abstraction and Indeterminate Forms of Limits

    \lim_{x\to 0} \frac{2x^2-3x-4}{x^2-1}

    I had this problem on a quiz today and I don't feel comfortable with how I approached it. Since the highest degree of both the variables in the numerator and denominator are the same, I assumed that the limit as x approaches zero would be the quotient of their coefficients, 2. My professor teaches this method of limit taking much more abstractly though when direct substitution doesn't apply, and that's where I am confused. He'll use notation like \frac{1}{\infty} or -\infty-\infty to find the limit of a problem where direct substitution doesn't apply.

    For the problem in this post I tried taking a difference of squares in the denominator and factoring the top but that didn't get me anywhere. I guess my question is whether or not anyone's familiar with the abstraction process for indeterminate forms; e.g., \frac{1}{0} being equal to infinity, and if the right answer here is 2?
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    Re: Abstraction and Indeterminate Forms of Limits

    which limit did you wish to know about?


    \lim_{x \to 0} \frac{2x^2-3x-4}{x^2-1} = 4

    \lim_{x \to \infty} \frac{2x^2-3x-4}{x^2-1} = 2
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    Re: Abstraction and Indeterminate Forms of Limits

    \lim_{x \to 0}.

    In this case, all that's necessary is direct substitution of zero into the x? I think I'm just looking at limits too hard.
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    Re: Abstraction and Indeterminate Forms of Limits

    direct substitution works in this particular case.
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    Re: Abstraction and Indeterminate Forms of Limits

    In a case where it doesn't, specifically for polynomials in rational functions, does that mean it's in an indeterminate form?
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    Re: Abstraction and Indeterminate Forms of Limits

    yes
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    Re: Abstraction and Indeterminate Forms of Limits

    Quote Originally Posted by AZach View Post
    In a case where it doesn't, specifically for polynomials in rational functions, does that mean it's in an indeterminate form?
    Also note that, if P(a)= 0, with P(x) a polynomial, x- a must be a factor. That means that if, setting x= a in both numerator and denominator, both numerator and denominator give 0, you can factor (x- a) out of each and cancel.
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