How would you find the end behavior of a radical like... (x^2+3)^1/2-x

We are asked to find the end behavior of the radical function f(x)=sqrt(x^2+3)-x .
The function has two terms; there is a radical expression and the linear polynomial -x.
Consider the radical expression. First, it is always positive (assuming this is a real-valued function.) At x = 0, it has value sqrt(3) . As x increases without bound, the 3 is irrelevant and the expression has a value that tends to |x|=x. Thus, as x increases without bound, f(x) tends to x-x=0. As x decreases without bound, the radical expression tends to |x|, so f(x) tends to infinity. There is also a slant asymptote, at f(x) = -2x.
Another way to see this is to consider the graph:


We can also use limits. Rationalize the numerator:
(sqrt(x^2+3)-x)*(sqrt(x^2+3)+x)/(sqrt(x^2+3)+x)
=(x^2+3-x^2)/(sqrt(x^2+3)+x)=3/(sqrt(x^2+3)+x)
For x>0: lim_(x->oo)3/(sqrt(x^2+3)+x)=lim_(x->oo)3/(sqrt(x^2+3)+x)*(1/x)/(1/sqrt(x^2))
=lim_(x->oo)(3/x)/(sqrt(1+3/x^2)+1)=0
For x<0 : note that sqrt(x^2)=|x|=-x so we get:
=lim_(x->-oo)(3/x)/(sqrt(1+3/x^2)-1)=lim_(x->-oo)(3/x)/0=oo
http://mathworld.wolfram.com/Limit.html