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Prove:

$\displaystyle \lim_{x\to 1} \sqrt{x} = 1$

$\displaystyle |\sqrt{x} - 1| < \epsilon \space \text{such that}\space |x - 1| < \delta$

Let $|x - 1| < 1 \implies |x| < 2 \implies -2 < x < 2$

$\sqrt{x} - 1 \implies \sqrt{x} - 1 < \sqrt{2} - 1$

$|\sqrt{x} - 1| < \sqrt{2} -1 $

$-2 < x < 2$

$ -3 < x - 1 < 1 \implies 1 < |x - 1| < 3$

$1 > \sqrt{2} - 1$

$\therefore |\sqrt{x} - 1| < \sqrt{2} - 1 < 1 < |x - 1| < \delta$

Finally, $|\sqrt{x} - 1| < |x - 1| < \delta$

Therefore, $\delta = \min(1, \epsilon)$ $\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space \blacksquare $

Amad27
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  • Its supposed to be a proof. – Amad27 Nov 13 '14 at 15:29
  • Amad27, what you've shown here is the scratchwork that precedes the writing of a proof. Take a look at the answer by Clarinetist at http://math.stackexchange.com/questions/418961/epsilon-delta-proof-that-lim-limits-x-to-1-frac1x-1 -- in particular, the distinction he makes between scratchwork and proof. At the same time, don't despair. Learning to write correct proofs takes time, and everyone struggles with it at first. I know I did. – Barry Cipra Nov 13 '14 at 15:39

3 Answers3

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The elements of a proof are there, but the logic is not clear.

We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that if $|x-1|\lt \delta$ then $|\sqrt{x}-1|\lt \epsilon$.

Let $\epsilon\gt 0$ be given. Let $\delta=\min(1,\epsilon)$. If $|x-1|\lt \delta$, then $x\gt 0$, and $$|\sqrt{x}-1|=\left|\frac{x-1}{\sqrt{x}+1}\right|\lt |x-1|\lt \epsilon.$$

André Nicolas
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  • so mine wont be considered correct right? – Amad27 Nov 13 '14 at 15:27
  • Your idea is basically right, so a reasonable grader would assign fairly good part marks. But because of a lack of clarity, you would not get full marks. Already the third line, the one that starts with $|\sqrt{x}=1|\lt \epsilon$ would cause a loss of marks. And there are some things that are just plain wrong, like (a few lines down) $1\lt |x-1|\lt 3$. – André Nicolas Nov 13 '14 at 15:36
  • Why is the last statement wrong? The $1 < |x - 1| < 3$? – Amad27 Nov 13 '14 at 15:37
  • $x$ is actually chosen to be close to $1$, which $|x-1|\gt 1$ contradicts. – André Nicolas Nov 13 '14 at 15:38
  • I see. What is the work around then? – Amad27 Nov 13 '14 at 15:40
  • I have tried in my solution to use what may be the geometric idea you had, to conclude that if $\delta$ is less than $1$, then $|\sqrt{x}-1|\lt |x-1|$. – André Nicolas Nov 13 '14 at 15:44
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As André said, make your proof a bit clearer and it will work. Since you haven't expressed the necessity of introducing $\delta$, you might want to consider the following simpler alternative: $$\displaystyle \left\lvert \sqrt{x}-1 \right\rvert < \epsilon \\-\epsilon<\sqrt{x}-1<\epsilon \\ (1-\epsilon)^2<x<(1+\epsilon)^2 \\ 1-2\epsilon+\epsilon^2<x<1+2\epsilon+\epsilon^2.$$

Vincenzo Oliva
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In this case you can directly solve the inequation $$ |\sqrt{x}-1|<\varepsilon $$ where $\varepsilon>0$. The inequation is equivalent to $$ \begin{cases} \sqrt{x}<1+\varepsilon\\ \sqrt{x}>1-\varepsilon \end{cases} $$ If $\varepsilon>1$, the solution set contains the interval $[0,(1+\varepsilon)^2)$ which is a neighborhood of $1$.

If $0<\varepsilon\le1$, the system of inequations is equivalent to $$ (1-\varepsilon)^2<x<(1+\varepsilon)^2 $$ which is again a neighborhood of $1$.

If you absolutely need a $\delta>0$, you can take $\delta=1$ for $\varepsilon>1$ and $$ \delta=\min\{(1+\varepsilon)^2-1,1-(1-\varepsilon)^2\} $$ for $0<\varepsilon\le1$. Since $(1+\varepsilon)^2-1=2\varepsilon+\varepsilon^2$ and $1-(1-\varepsilon)^2=2\varepsilon-\varepsilon^2$, you see that $\delta=2\varepsilon-\varepsilon^2$ is a good choice.

egreg
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  • but what about just $\delta = \epsilon$ like in mine? – Amad27 Nov 13 '14 at 15:35
  • @Amad27 Nothing bad at all! What I found is the largest value for $\delta$ given $\varepsilon>0$ (which is not needed, actually, but I know that some teachers are fussy about this). Your proof is very badly exposed, by the way. – egreg Nov 13 '14 at 15:37
  • thanks. But as Andre Nicolas pointed, Is the statement $1 < |x - 1| < 3$ wrong? – Amad27 Nov 13 '14 at 15:38
  • @Amad27 That's one of the several weaknesses in your proof. You can assume $1<|x-1|<3$, because this has no consequence on the verification of the limit. – egreg Nov 13 '14 at 15:40
  • Right, we are allowed to make assumptions. The definition only asks for one $\delta$, if we make an assumption then thats safe isnt it? – Amad27 Nov 13 '14 at 15:41
  • @Amad27 Yes, so long as you restrict $x$ to vary in a neighborhood of $1$ (excluding $1$, of course). – egreg Nov 13 '14 at 15:42
  • So then the proof is correct, but I just wrote it in a terrible way? – Amad27 Nov 13 '14 at 15:44
  • @Amad27 More or less, that's how I would grade it. – egreg Nov 13 '14 at 15:45