# Evalute the following limit

1. Nov 16, 2013

Hey guys, I have been trying this question for 40 minutes and I can't get the answer.
Evaluate lim as x approaches positive infinity(sqrt(x^2+1)-x).
I tried the La-Hopital's rule, but even then I still get inf/inf. I know there is an algebraic manipulation I need to make, but I keep getting it wrong.

2. Nov 16, 2013

### hilbert2

I'd suggest you show that if $f(x)=\sqrt{x^{2}+1}-x$, then $0 \leq f(x) \leq \sqrt{x^{2}}+\frac{1}{x}-x$ for positive $x$, and because $\lim_{x \to \infty}\left(\sqrt{x^{2}}+\frac{1}{x}-x\right)=0$, the limit of $f(x)$ must be zero when $x$ approaches infinity.

You can prove that $f(x) \leq \sqrt{x^{2}}+\frac{1}{x}-x$ by assuming the contrary and deriving a contradiction.

3. Nov 16, 2013

Thanks hilbert2,
I see what you are saying. I also just found another way to do it, ill post it:
Multiplying the expression by (sqrt(x^2+1) + x)/(sqrt(x^2+1)+x) you get lim(1/(sqrt(x^2+1)+x) = 0.
I see that you have used the sandwich theorem, and i suppose that is an easier way to do it. I am confused on how you got your upper limit 1/x, could you please explain how you got that upper limit. Thanks

4. Nov 16, 2013

Ah, did you just arbitrarily pick the upper limit which is larger than f(x)?

5. Nov 16, 2013

### hilbert2

Let's suppose the contrary, that $\sqrt{x^{2}+1}>\sqrt{x^{2}}+\frac{1}{x}$. As we can suppose $x$ is positive, we can square both sides of the inequality to get
$x^{2}+1>x^{2}+2\frac{|x|}{x}+\frac{1}{x^{2}}$. When $x$ is positive, $\frac{|x|}{x}=1$, and therefore we get $1>2+\frac{1}{x^{2}}$, which is a contradiction.

You can choose any upper limit that approaches zero when $x \to \infty$.