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.

  • $\begingroup$ Please edit your question to show the work you have done. $\endgroup$ – N. F. Taussig Mar 11 '15 at 12:43

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 $N=\frac1{\sqrt2\epsilon}$.

  • 1
    $\begingroup$ I think you should justify why $x^{2} + 1 + x\sqrt{x^{2} + 1} \geq 2x^{2}$. It's because $\sqrt{x^{2} + 1} \geq \sqrt{x^{2}} = x \implies x^{2} + 1 + x \sqrt{x^{2} + 1} \geq x^{2} + 1 + x^{2} \geq 2x^{2} + 1 \geq 2x^{2}$. $\endgroup$ – layman Mar 11 '15 at 12:50
  • $\begingroup$ I think that I shouldn't justify it and I should not give a complete answer. $\endgroup$ – user63181 Mar 11 '15 at 12:57
  • $\begingroup$ Ok. Not that it makes a huge deal, but really we should have $x > \dfrac{1}{\sqrt{2 \epsilon}}$, right? $\endgroup$ – layman Mar 11 '15 at 13:00
  • $\begingroup$ In my answer I just said it's sufficient to take $x>\frac1{\epsilon\sqrt2}$ so it isn't a necessary condition.@user46944 $\endgroup$ – user63181 Mar 11 '15 at 13:05

\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.


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.


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