Another limit task, I've multiplied by the conjugate, now what? The task: 
$\lim_{x\to\infty} \sqrt{x^2+1} -x $
I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this
$\lim_{x\to\infty} \frac{1}{\sqrt{x^2+1} +x} $
Is this correct so far? And what would be the next step? 
 A: Remember that $a^{2}-b^{2}=(a-b)(a+b)$, put $a=\sqrt{x^{2}+1}$ and $b=x$ get:
$$(\sqrt{x^{2}+1}+x)(\sqrt{x^{2}+1}-x)=(\sqrt{x^{2}+1})^{2}-(x)^{2} = x^{2}+1-x^{2}= 1.$$
The end...
A: The multiplication of conjugates is not correct: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)=(x^2+1)-x^2=1$
A: We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$
A: HINT: What you have to do is this : $$ \sqrt{x^{2}+1} -x \times \frac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1} + x} = \frac{1}{\sqrt{x^{2}+1} +x}$$ then proceed to evaluate your limit.
Now observe that $\lim_{x \to \infty} \frac{1}{x} = 0$. 
A: $$\underset{x\rightarrow \infty }{\lim }\sqrt{x^{2}+1}-x=\underset{%
x\rightarrow \infty }{\lim }\frac{\left( \sqrt{x^{2}+1}-x\right) \left( 
\sqrt{x^{2}+1}+x\right) }{\sqrt{x^{2}+1}+x}=\underset{x\rightarrow \infty }{%
\lim }\frac{1}{\sqrt{x^{2}+1}+x}$$

Added
$$\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}=\infty$$
and (see Wikipedia, limits involving infinity and properties) 
$$\underset{x\rightarrow \infty }{\lim }\frac{1}{\sqrt{x^{2}+1}+x}=\dfrac{\underset{x\rightarrow \infty }{\lim }1}{\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}}=\dfrac{1}{\infty}=0$$
A: HINT $\rm\ \ \ \ $ Put $\rm\ f = x,\ g = 1\ $ in the following simple
LEMMA $\rm\displaystyle\ \ \ \sqrt{f^2+g}\ -\ f\ \to\ 0 \ \ \ if\ \ \ \frac{f}g \to\ \infty\ \ \ and\ \ \frac{1}g\ \to\ \infty\ \ or\ \ r\in \mathbb R$
Proof $\rm\displaystyle\quad\quad\ \ \sqrt{f^2+g}\ -\ f\ \ =\ \frac{g}{\sqrt{f^2+g}\ +\ f}\ =\ \frac{1}{\sqrt{(\frac{f}g)^2+\frac{1}g}\ +\ \frac{f}g}$
