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.

\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.
