Is negative infinity divided by infinity still indeterminate?

[Glamis321]

Just as the title states, I'm working on a problem and have come to negative infinity divided by infinity. Is this an indeterminate form? I know that if they are both positive it is indeterminate, but I can't remember if one being negative makes a difference.

 

[fzero]

You should post the expression in question. The answer is that it depends and might require more care in taking a limit. For example

$$\lim_{r\rightarrow\infty} \frac{-r^2}{r^2} = \frac{-\lim_{r\rightarrow\infty} r^2}{\lim_{r\rightarrow\infty} r^2}$$

has indeterminate numerator and denominator, but the ratio is actually finite. However

$$\lim_{r\rightarrow\infty} \frac{-r^3}{r^2} = \frac{-\lim_{r\rightarrow\infty} r^3}{\lim_{r\rightarrow\infty} r^2}$$

is indeterminate.

 

[Mark44]

Yes, [-infty/infty] is an indeterminate form, just as [infty/infty] is.

 

[Glamis321]

I'll give it a shot but I don't know how to use the latex codes.

The original problem was Lim (2x – square root of [4x²+x]),
x->∞

Just by plugging in the ∞, I came up with ∞-∞, which I know is indeterminate. So I multiplied in the conjugate of that function and came up with this.


-x divided by (2x+ sqrt of [4x²+x])

 

[Mark44]

Yes. So far, so good. Here it is in LaTeX. Click the expression to see what I did.
$$\lim_{x \to \infty} \frac{-x}{2x + \sqrt{4x^2 + x}}$$

You can factor x² out of both terms in the radical, bringing out a factor of x, which means you can factor x out of the two terms in the denominator.