What is the correct upper limit to evaluate the given limit?

[baddin]

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.
Thanks in advance.

 

[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.

 

[baddin]

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

 

[baddin]

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

 

[hilbert2]

I am confused on how you got your upper limit 1/x, could you please explain how you got that upper limit. Thanks

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 $x \to \infty$.