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Math Help - Limits At Infinity With Radical

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    Limits At Infinity With Radical

    I am reading an example about evaluating limits at infinity. The function in the example is f(x) = \frac{3x -2}{\sqrt{2x^2 + 1}}. What I am confused about is why do that put a negative sign in front of the radical when they are evaluating the limit of the function at negative infinity?
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    Re: Limits At Infinity With Radical

    Quote Originally Posted by Bashyboy View Post
    I am reading an example about evaluating limits at infinity. The function in the example is f(x) = \frac{3x -2}{\sqrt{2x^2 + 1}}. What I am confused about is why do that put a negative sign in front of the radical when they are evaluating the limit of the function at negative infinity?
    Is this the question, {\lim _{x \to  - \infty }}\frac{3x -2}{\sqrt{2x^2 + 1}}=-\frac{3}{\sqrt2}~?

    If it is then note that for x<0 the numerator is negative and the denominator is positive.
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    Re: Limits At Infinity With Radical

    Yes, that's what I was asking. I was just wondering if there was a reason for that.
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    Re: Limits At Infinity With Radical

    Quote Originally Posted by Bashyboy View Post
    Yes, that's what I was asking. I was just wondering if there was a reason for that.
    I gave you the answer. Read the reply again.
    You have \frac{-}{+}=-
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