2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Please edit your question to show the work you have done. $\endgroup$ – N. F. Taussig Mar 11 '15 at 12:43
5
$\begingroup$

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

$\endgroup$
  • 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
1
$\begingroup$

\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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.