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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 $\epsilon-\delta$)?

Martin Sleziak
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Gabriel Bianconi
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3 Answers3

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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 $\delta \le 1$ ..."

Emanuele Paolini
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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.

awllower
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  • The backward approach. :D – awllower Jul 23 '13 at 15:57
  • That's an interesting approach! Thanks. – Gabriel Bianconi Jul 23 '13 at 16:05
  • @awllower How $0\lt\delta\lt \min{|-2\epsilon+\epsilon^2|,|2\epsilon+\epsilon^2|}$ proves that $|x-1|<\delta$? Please elaborate this step. – user1942348 Mar 12 '17 at 12:07
  • 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. – awllower Mar 12 '17 at 14:13
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    You need to let $\epsilon\le 1$ to square/raise to the even positive power $2$ the inequality. E.g., $-4<3$, but $(-4)^2\not < 3^2$. – user236182 Nov 02 '17 at 18:38
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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.

ncmathsadist
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math4fun
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  • You might want to review: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference?rq=1 – Amzoti Nov 16 '13 at 02:03