Real Analysis - Prove limit $\lim_{x\to 25} \sqrt x = 5$

Question

Prove that $\lim\limits_{x\to 25} \sqrt x = 5$ using $\epsilon$ and $\delta$.

can someone check my work please?

$$|x-a|<\delta \qquad\Rightarrow\qquad |f(x)-L|<\varepsilon$$ To prove this we must show the following:
There exists a $\delta>0$ for every $\varepsilon>0$ such that $$|x-25|<\delta \qquad\Rightarrow\qquad |\sqrt x-25|<\varepsilon.$$ \begin{gather*} -\varepsilon < \sqrt x - 25 < \varepsilon \\ 25-\varepsilon < \sqrt x < \varepsilon+25 \\ (25-\varepsilon)^2 < x < (\varepsilon+25)^2 \\ 25-\varepsilon < \sqrt x < \varepsilon+25 \\ |\sqrt x-25| < \varepsilon=\delta \end{gather*} Let $\varepsilon>0$ be given. Then choose $\varepsilon=\delta$. \begin{gather*} -\varepsilon < x-25 < \varepsilon \\ 25-\varepsilon < x < 25+\varepsilon \\ -\varepsilon < x-25 < \varepsilon \\ -\sqrt\varepsilon < \sqrt{x}-\sqrt{25} < \sqrt\varepsilon \\ -\varepsilon < \sqrt{x}-\sqrt{25} < \varepsilon \\ |\sqrt{x}-\sqrt{25}|<\varepsilon \\ \lim_{x\to25} \sqrt x = 5 \end{gather*} Hence proved.

Answer 1

There rae lot of errors in your work.

For all $x$ such that $|x-25| < 9$ we have $16 < x < 34 $ then $4< \sqrt x$ then $9 < \sqrt x + 5$ and $\frac{1}{\sqrt x + 5} < \frac 19$

$|\sqrt x- 5|=\frac{|x-25|}{\sqrt x+5} \leq \frac19|x-25|$

For all $\varepsilon > 0$, let $\delta=\min(9,9\varepsilon)$

It's easy to show that $$\forall x \in \Bbb R,\quad |x-25| < \delta \Rightarrow |\sqrt x - 5| < \varepsilon$$

Answer 2

Though your development is unclear (and there are typos), the answer is correct.

Given an $\epsilon$, you need to find a $\delta$ such that

$$|x-25|<\delta\implies|\sqrt x-5|<\epsilon.$$

You can factor and rewrite

$$|\sqrt x-5|<\frac\delta{\sqrt x+5}\implies|\sqrt x-5|<\epsilon.$$

Hence $\delta\le(\sqrt x+5)\epsilon$ establishes the inequality for any $\epsilon$, and $\delta=5\epsilon$ or $\delta=\epsilon$, which are tighter values, are also valid.

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The "hard way" is

$$|\sqrt x-5|<\epsilon,$$ $$5-\epsilon<\sqrt x<5+\epsilon,$$ $$(5-\epsilon)^2-25<x-25<(5+\epsilon)^2-25,$$ $$|x-25|<\min(|(5-\epsilon)^2-25|,(5+\epsilon)^2-25)\\ =\epsilon\min(|\epsilon-10|,\epsilon+10).$$

Note that the squaring from the second to the third line requires $\epsilon\le5$ so that $\delta\le5\epsilon$ can be used.

Answer 3

You seem to have assumed $\sqrt{a + b} = \sqrt a + \sqrt b$ which is of course not true.

To continue your method.

$|\sqrt{x} - 5| < \epsilon$

$-\epsilon < \sqrt x - 5 < \epsilon$

$5 - \epsilon < \sqrt{x} < 5 + \epsilon$

$(5-\epsilon)^2 < x < (5 + \epsilon)^2$

$25 - 10\epsilon+ \epsilon^2 < x < 25 + 10\epsilon + \epsilon^2$

$-10\epsilon + \epsilon^2 < x -25 < 10\epsilon + \epsilon^2$.

So we need some $\delta$ so that $-10\epsilon + \epsilon^2 \le -\delta < x-25 <\delta \le 10\epsilon + \epsilon^2$

If $\epsilon < 1$ then $\epsilon^2 < \epsilon$ so

$-10\epsilon + \epsilon^2 < -10\epsilon + \epsilon = -9\epsilon$

and $10 \epsilon + \epsilon^2 > 10\epsilon > 9\epsilon$

So $\delta = \min(9\epsilon, 9)$ will be such a $\delta$.

So our proof:

Case 1: If $\epsilon < 1$ then if $\delta = 9\epsilon$

$|x - 25| < \delta \implies$

$-\delta < x - 25 < \delta \implies$

$25 - \delta < x < 25 + \delta$

$25 - 9\epsilon < x < 25 + 9\epsilon < 25 + 10\epsilon + \epsilon^2$

$25 - 10\epsilon + \epsilon^2 < 25-9\epsilon < x < 25+10\epsilon + \epsilon^2$

$0 < 4^2 < (5 - \epsilon)^2 < x < (5+\epsilon)^2$

$5 - \epsilon < \sqrt{x} < 5 + \epsilon$

$|\sqrt{x} - 5| < \epsilon$.

Case 2:

$\epsilon \ge 1$ then

$|25 - x| < 9 \implies$

$16 < x < 35$

$4 <\sqrt{ x} < \sqrt{35} < 6$

$-\epsilon \le -1 < \sqrt{x} - 5 < 1 \le \epsilon$

$|\sqrt{x} -5| < \epsilon$.

So for any $\epsilon > 0$ there is a $\delta$ so that $|x - 25|< \delta \implies |\sqrt{x} - 5| < \epsilon$ so $\lim_{x\rightarrow 25} \sqrt{x} = 5$.

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Although Caridini's method is more clever and efficient. However if you didn't see it... well, you can always use brute force as I did.