Limit of a function: $\lim_{x\to \infty} \frac x{\sqrt{1+x^2}}$ I'm having trouble finding $\lim_{x\to \infty} {x\over \sqrt{1+x^2}}$ by using the formal analysis proof i.e. $\forall \epsilon>0$ $\exists N: \left|f(x)-L\right|<\epsilon$ $\forall x>N$. I know the answer is 1 but I can't manage to prove it sufficiently. The main problem is rearranging $\left|f(x)-1\right|$ to get a nice equation of $x$ in terms of $\epsilon$. Some help would be greatly appreciated.
So far I have got to $\left|\frac{x}{\sqrt{1+x^2}}-1\right|<\epsilon$ and now I need $x$ to be in terms of $\epsilon$ in order to find a suitable $N$ for $x$ to be greater than.
 A: We have
$$|f(x)-1|=\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}}=\frac1{\sqrt{x^2+1}(\sqrt{x^2+1}+x)}\le\frac{1}{2x^2}<\epsilon\iff x>\frac1{\sqrt2\epsilon}$$
so it suffices to take $N=\frac1{\sqrt2\epsilon}$.
A: \begin{split}
&\left|\frac{x}{\sqrt{1+x^2}}-1\right|\\
=&\left|\frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}}\right|\\
=&\left|\frac{(x-\sqrt{1+x^2})(x+\sqrt{1+x^2})}{\sqrt{1+x^2}(x+\sqrt{1+x^2})}\right|\\
=&\left|\frac{1}{1+x^2+x\sqrt{1+x^2}}\right|\\
<&\frac{1}{x^2}
\end{split}
Then we can say that $$\forall \varepsilon>0, \exists N>\sqrt{\frac{1}{\varepsilon}}>0, \forall x>N, |f(x)-1|<\frac{1}{x^2}<\frac{1}{N^2}<\varepsilon$$
Hope this can help you.
A: Yet another observation.
Let $\varepsilon > 0.$ We have
$$|\frac{x}{\sqrt{1 + x^{2}}} - 1| = |\frac{x - \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}}| \leq |x - \sqrt{1 + x^{2}}| \leq \frac{1}{x + \sqrt{1 + x^{2}}} < \frac{1}{x} < \frac{1}{N},$$
where we have used the fact that $\sqrt{1 + x^{2}} \geq 1$ and the multiplication factor $\frac{x + \sqrt{1 + x^{2}}}{x + \sqrt{1 + x^{2}}}.$ 
Thus taking $N := 1/\varepsilon$ finishes the proof. 
