For this limit $\lim_{x \to 5}{\sqrt{x-1}}=2$ find $ δ$

Question

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.

Answer 1

$\newcommand{\R}{\mathbb{R}}$ 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 $0 < \delta \le 1 \cdot\sqrt{4} = 2$.

Answer 2

$$\left\vert\sqrt{x-1}-2\right\vert<1\\ \Leftrightarrow -1<\sqrt{x-1}-2<1\\ \Leftrightarrow 1<\sqrt{x-1}<3\\ \Leftrightarrow 1<x-1<9\\ \Leftrightarrow -3<x-5<5\\$$ Thus, any $\delta\le 3$ works for $\varepsilon=1$.

Answer 3

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.

Answer 4

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..