1. ## Limits At Infinity With Radical

I am reading an example about evaluating limits at infinity. The function in the example is $\displaystyle f(x) = \frac{3x -2}{\sqrt{2x^2 + 1}}$. 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?

2. ## Re: Limits At Infinity With Radical

Originally Posted by Bashyboy
I am reading an example about evaluating limits at infinity. The function in the example is $\displaystyle f(x) = \frac{3x -2}{\sqrt{2x^2 + 1}}$. 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, $\displaystyle {\lim _{x \to - \infty }}\frac{3x -2}{\sqrt{2x^2 + 1}}=-\frac{3}{\sqrt2}~?$

If it is then note that for $\displaystyle x<0$ the numerator is negative and the denominator is positive.

3. ## Re: Limits At Infinity With Radical

Yes, that's what I was asking. I was just wondering if there was a reason for that.

4. ## Re: Limits At Infinity With Radical

Originally Posted by Bashyboy
Yes, that's what I was asking. I was just wondering if there was a reason for that.
You have $\displaystyle \frac{-}{+}=-$