11
$\begingroup$

I'm trying to teach myself how to do $\epsilon$-$\delta$ proofs and would like to know if I solved this proof correctly. The answer given (Spivak, but in the solutions book) was very different.


Exercise: Prove $\lim_{x \to 1} \sqrt{x} = 1$ using $\epsilon$-$\delta$.

My Proof:

We have that $0 < |x-1| < \delta $.

Also, $|x - 1| = \bigl|(\sqrt{x}-1)(\sqrt{x}+1)\bigr| = |\sqrt{x}-1||\sqrt{x}+1| < \delta$.

$\therefore |\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|}$

Now we let $\delta = 1$. Then \begin{array}{l} -1<x-1<1 \\ \therefore 0 < x < 2 \\ \therefore 1 < \sqrt{x} + 1<\sqrt{2} + 1 \\ \therefore \frac{1}{\sqrt{x} + 1}<1. \end{array}

We had that $$|\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|} \therefore |\sqrt{x}-1|<\delta$$

By letting $\delta=\min(1, \epsilon)$, we get that $|\sqrt{x}-1|<\epsilon$ if $0 < |x-1| < \delta $.

Thus, $\lim_{x \to 1} \sqrt{x} = 1$.


Is my proof correct? Is there a better way to do it (still using $\epsilon-\delta$)?

$\endgroup$
  • 7
    $\begingroup$ Looks good to me. Although perhaps unnecessary because it's easy to argue that $\frac{1}{\sqrt{x}+1} \le 1$. $\endgroup$ – Cameron Williams Jul 23 '13 at 15:46
  • $\begingroup$ @CameronWilliams I see. In that case, do I still need to let $\delta=min(1, \epsilon)$ or can it simply be $\delta=\epsilon$? Thank you. $\endgroup$ – Gabriel Bianconi Jul 23 '13 at 15:55
  • 2
    $\begingroup$ You can simply let $\varepsilon = \delta$. $\endgroup$ – Cameron Williams Jul 23 '13 at 16:00
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Jul 23 '13 at 16:00
  • $\begingroup$ @CameronWilliams Thanks! $\endgroup$ – Gabriel Bianconi Jul 23 '13 at 16:07
4
$\begingroup$

The proof is correct but can be simplified. You don't need the part "Now let $\delta=1$...". In fact it is always true that $$ \frac{1}{\sqrt x + 1} \le 1 $$ since $\sqrt x \ge 0$.

Also, a matter of style. In the first line you don't have $0 < |x-1|<\delta$ but you suppose it (this is because $\delta$ is not already been given, but has to be found yet). The same when you write "let $\delta = 1$" you should write "if $\delta \le 1$ ..."

$\endgroup$
  • $\begingroup$ Thank you for the suggestions! $\endgroup$ – Gabriel Bianconi Jul 23 '13 at 16:03
  • $\begingroup$ Wrong. You need to let $\delta\le 1$ because if $\delta>1$, then $\sqrt{x}$ may not be a real number. $\endgroup$ – user236182 Nov 2 '17 at 18:21
3
$\begingroup$

Your proof is correct.
We can also adopt the following:
Since $|\sqrt x-1|\lt \epsilon$ is equivalent with $1-2\epsilon+\epsilon^2\lt x\lt 1+2\epsilon+\epsilon^2$, we can choose $\delta$ so that $0\lt\delta\lt \min\{|-2\epsilon+\epsilon^2|,|2\epsilon+\epsilon^2|\}$.
Hope this helps.

$\endgroup$
  • $\begingroup$ The backward approach. :D $\endgroup$ – awllower Jul 23 '13 at 15:57
  • $\begingroup$ That's an interesting approach! Thanks. $\endgroup$ – Gabriel Bianconi Jul 23 '13 at 16:05
  • $\begingroup$ @awllower How $0\lt\delta\lt \min\{|-2\epsilon+\epsilon^2|,|2\epsilon+\epsilon^2|\}$ proves that $|x-1|<\delta$? Please elaborate this step. $\endgroup$ – user1942348 Mar 12 '17 at 12:07
  • $\begingroup$ Our aim here is that for every $|x-1|\lt\delta$, we have $|\sqrt x-1|\lt\varepsilon$. Now if $0\lt\delta\lt \min\{|-2\varepsilon+\varepsilon^2|,|2\varepsilon+\varepsilon^2|\}$, then for $|x-1|\lt\delta$, we deduce $1-\delta\lt x\lt1+\delta$, from which the conclusion follows. Note also that when $\varepsilon$ is small, $\varepsilon^2-2\varepsilon$ is negative. $\endgroup$ – awllower Mar 12 '17 at 14:13
  • 1
    $\begingroup$ You need to let $\epsilon\le 1$ to square/raise to the even positive power $2$ the inequality. E.g., $-4<3$, but $(-4)^2\not < 3^2$. $\endgroup$ – user236182 Nov 2 '17 at 18:38
0
$\begingroup$

Using your work here is another flavor of this proof:

Let $\epsilon >0$, and put $\delta= \epsilon(\sqrt{x}+1)$.

Assume $0<|x−1|<\delta$.

Then $$|F(x)−L|=|x−1| =∣(\sqrt{x}−1)(\sqrt{x}+1)∣.$$ By our assumption that $0<|x−1|<\delta$, we have $$|F(x) - L| <|1/(\sqrt{x}+1)|\delta = (1/(\sqrt{x}+1))ϵ(\sqrt{x}+1)) = \epsilon.$$

By letting $\delta=\epsilon(\sqrt{x}+1)$, we get that $|x−1|<\epsilon$ if 0<|x−1|<δ. Thus, $\lim_{x\to 1} F(x) = L$

The scratch work is usually omitted as far as finding the co-efficient of delta itself. Then just find the co-eff inverse and include epsilon and it falls out at the end. There is a good link on math exchange that shows a template of how to structure delta-epsilon proofs. It is how I learned to write them up, and this is that method. Good luck.

P.S. * represents multiplication, and the square roots were not working. I am unfamiliar with LaTex so this is the best I can do.

$\endgroup$

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.