# Thread: Limits and negative infinity

1. ## Limits and negative infinity

Hello, my name is Erik and I知 looking for help regarding an explanation (or a proof) of why the following concept is true. (My teacher cannot give me one and neither can the calculus manual, they both simply state that this is just how it is)

problem:
img508.imageshack.us/img508/4498/problemap1.gif

I知 not sure I知 using the correct notation in solving this problem but really what I知 looking for is the proof or explanation why this is possible and why this is how it should be solved.

Thanks

Erik

2. Are you wondering why it is divided by -x?.

It is helpful to manipulate the function so that powers of x become powers of 1/x.

This can be done by dividing the numerator and denominator by

$\sqrt{x^{2}}=|x|$

Since $x\rightarrow{-\infty}$, the values of x are eventually negative, so we can replace |x| by -x where desireable.

Does that help any?.

3. Originally Posted by galactus
Are you wondering why it is divided by -x?.

It is helpful to manipulate the function so that powers of x become powers of 1/x.

This can be done by dividing the numerator and denominator by

$\sqrt{x^{2}}=|x|$

Since $x\rightarrow{-\infty}$, the values of x are eventually negative, so we can replace |x| by -x where desireable.

Does that help any?.
I understand why we have to divide by -x my question is more or less why this can be done. How could i communicate that since x->-inf i can write x as -x? If im understanding this right i might as well write -x as x since it is just variables. But this does not work since if i write it as x i wont solve it properly.

Thanks alot

Erik

4. ## You can do this

Originally Posted by tyro89
I understand why we have to divide by -x my question is more or less why this can be done. How could i communicate that since x->-inf i can write x as -x? If im understanding this right i might as well write -x as x since it is just variables. But this does not work since if i write it as x i wont solve it properly.

Thanks alot

Erik
You can do this using the definition of |x|... $|x|=-x,\forall{x}<0$ and $|x|=x,\forall{x}>0$

5. Originally Posted by Mathstud28
You can do this using the definition of |x|... $|x|=-x,\forall{x}<0$ and $|x|=x,\forall{x}>0$
Thanks a lot i get it now