How to prove a function is positive or negative in $x \in \mathbb{R}$ A homework question:
I know the solution but I don't know how to prove that the function is negative XOR positive for $x \in \mathbb{R}$
f is continuous in $\mathbb{R}$.
$$\text{ prove that if } |f(x)| \ge x \text{ for } x \in \mathbb{R} \text { then: }
 \lim_{x\to\infty} f(x) = \infty \text{  or  } \lim_{x\to\infty} f(x) = -\infty$$
Now once I prove that the function is negative XOR positive it's relatively simple to prove that the limits are at infinity.
How do I prove that there is no $f(x) = 0$?
Thanks!

The function, as you stated it, is not exclusively negative nor exclusively positive.
There's a really simple counter example.  If f(x)=x for all x in R, then f is continuous everywhere, |f(x)|>=x everywhere, and even one of the limits is satisfied (limit as x goes-to infinity of f(x) is infinity).  But clearly, f(0)=0, f(-1)=-1, and f(1)=1.  So there is a point where f(x)=0, and f is not exclusively positive nor exclusively negative everywhere.
As to the most likely question you were probably asking, user6312 already answered it for you, but I'll type the same proof for completeness.  (Also, does anyone know where I can find a guide on how to get latex to work properly in here?  I can't get things like /mathbb{R} or /infinity or /in to work.  Maybe the rules have changed since I last used latex...)
If f(1)>=1, then f is positive for all x>1.  (Suppose there exists b>1 such that f(b)<1. Since |f(x)|>=x for all x, and b>1, f(b)<-b<0.  But if 10 and f(b)<0, there exists a point c, 1the coolest theorem ever.  Since c>0 and f(c)=0, =><=)
Since f(x) is positive for all x>1, then |f(x)|=f(x) for x>1.  Thus the limit x->infinity of f(x) is greater than limit x->infinity of x which is infinity.
If f(1)<=-1, then let g(x)=-f(x), and do the same proof, and you'll end with lim x->infinity f(x) = -infinity.
