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I want to verify the following limit using the definition of a limit:

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

I know that:

$$0<∣x-1∣<\delta \qquad\text{and}\qquad ∣\sqrt{x}-1∣<\varepsilon,$$

but then I do not have a clue on how to move on.

José Carlos Santos
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    Hint : $$\sqrt{x}-1=\frac{x-1}{\sqrt{x}+1}$$ – Peter Sep 08 '17 at 13:35
  • Are you sure that $|\sqrt{x}-1| < \varepsilon$? Or is this the inequality that you are trying to prove holds? Do you understand the definition of a limit? Or does your confusion lie elsewhere? – Xander Henderson Sep 08 '17 at 13:48
  • I am starting to understand most of it, however I do not understand how √−1 can become (x−1/(√x+1), how is the math done here? –  Sep 08 '17 at 13:53
  • @JavaProgrammer The third binomial formula shows $$(\sqrt{x}-1)(\sqrt{x}+1)=(\sqrt{x})^2-1=x-1$$ Divide by $\sqrt{x}+1$ to get my equation – Peter Sep 08 '17 at 13:54
  • https://math.stackexchange.com/questions/450410/epsilon-delta-proof-that-lim-x-to-1-sqrtx-1?rq=1 – Stand with Gaza Sep 08 '17 at 15:07

2 Answers2

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Since $x-1=\sqrt x^2-1^2=\bigl(\sqrt x+1\bigr)\bigl(\sqrt x-1\bigr)$, and since $\sqrt x+1>1$ when $x>0$, if you take any $\varepsilon>0$ and if you choose $\delta>0$ such that $\delta\leqslant\varepsilon$ and that $\delta\leqslant1$, then$$|x-1|<\delta\iff\bigl(\sqrt x+1\bigr)\bigl\lvert\sqrt x-1\bigr\rvert<\delta\implies\bigl\lvert\sqrt x-1\bigr\rvert<\delta\leqslant\varepsilon.$$

José Carlos Santos
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$∣\sqrt{x}-1∣<\varepsilon$ means $\sqrt{x}-1<\varepsilon\land \sqrt{x}-1>-\varepsilon$

That is, for $0<\varepsilon\leq 1$

$\sqrt x < \varepsilon+1$ or $x < \varepsilon^2+2\varepsilon+1$ and

$\sqrt x > 1 -\varepsilon$ or $0<\varepsilon\leq 1 \land x>1-2\varepsilon+\varepsilon^2$

Thus we have $1-2\varepsilon+\varepsilon^2<x < \varepsilon^2+2\varepsilon+1$

which can be written as

$\varepsilon^2-2\varepsilon<x-1 < \varepsilon^2+2\varepsilon$

so taking $\delta=|\varepsilon^2-2\varepsilon|$

we are sure that if $|x-1|<\delta$ then $|\sqrt x -1| < \varepsilon$

For instance if $\varepsilon=0.001$ taking $\delta=|10^{-6}- 2 \times 10^{-3}|\approx 0.002$

if $|x-1|<0.002$ let's say $x=0.999$ then $|\sqrt x-1|\approx 0.0005<\varepsilon$

Raffaele
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