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

## Expert Answers 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 `

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