For this limit $\lim_{x \to 5}{\sqrt{x-1}}=2$ find $ δ$ Question on delta epsilon definition of a limit
For the limit  $\lim_{x \to 5}{\sqrt{x-1}}=2$, find a $δ > 0$ that works for $ε = 1$.
I'm getting $δ=1$, But I'm not 100% confirm that my ans is correct. Please help in this question.
 A: $\newcommand{\R}{\mathbb{R}}$
I will show that for all $ a > 0$,
$$
\lim_{x \to a} \sqrt{x}
=
\sqrt{a}
$$
We have to show that for any $\epsilon > 0$, there is $\delta > 0$ such that for all $x \ge 0$,
$$
|x - a| < \delta \Rightarrow |\sqrt{x} - \sqrt{a}| < \epsilon
$$
Let $|x - a| < \delta$. Then,
$$
|\sqrt{x} - \sqrt{a}|
=
\left|
\dfrac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{\sqrt{x} + \sqrt{a}}
\right|
=
\dfrac{|x-a|}{\sqrt{x} +\sqrt{a}}
<
\dfrac{\delta}{\sqrt{a}}
$$
Choose $\epsilon \ge \delta/\sqrt{a}$ or equivalently $\delta \le \epsilon \sqrt{a}$.
In your question, you can apply the change of variable $x - 1 = y$ to have exactly the same form I introduced, where $a = 4$. The change of variable (shifting) does not alter the dependence of $\delta$ over $\epsilon$.
Therefore, you can choose whatever $0 < \delta \le 1 \cdot\sqrt{4} = 2$.
A: $$\left\vert\sqrt{x-1}-2\right\vert<1\\
\Leftrightarrow -1<\sqrt{x-1}-2<1\\
\Leftrightarrow 1<\sqrt{x-1}<3\\
\Leftrightarrow 1<x-1<9\\
\Leftrightarrow -3<x-5<5\\$$
Thus, any $\delta\le 3$ works for $\varepsilon=1$.
A: Uniform Continuity can be used.
For $x,a>2$: Let $-\delta<x-a=(\sqrt{x-1}-\sqrt{a-1})(\sqrt{x-1}+\sqrt{a-1})<\delta$
So $\frac{-\delta}{\sqrt{x-1}+\sqrt{a-1}}<\sqrt{x-1}-\sqrt{a-1}<\frac{\delta}{\sqrt{x-1}+\sqrt{a-1}}$
The expression on the right achieves its maximum value ($\delta/2$) at x=2 and is less than that on the rest of the interval, so if $\delta\le\epsilon$ the expression on the right is always less than $\epsilon$.
$a=5$ is in the desired domain, so $\delta=\epsilon=1$ is acceptable.
A: We have
| root(x-1) -2|<1
Which implies
1< root(x-1) < 3
which on further solving we get
-3 < x-5 < 5
Which means that
0<|x-5|<5 comparing this with 0<|x-5|<delta
Hence delta is less than or equal to 5?
Plz correct me where i am wrong..
