Epsilon delta definition with restricted epsilon

Question

Here, I tried to prove:

$\lim\limits_{x \to 4} \sqrt{x} = 2$

$2 -\epsilon < \sqrt{x} < 2 + \epsilon$

Edit:

Mu Prime Math has told me that before squaring both sides of the inequality must be positive:$2 - \epsilon \geq 0$

$2 \geq \epsilon$

$\epsilon>0$

$0 < \epsilon \leq 2$

End of edit

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

$4- 4\epsilon +\epsilon^2 < x < 4 + 4\epsilon + \epsilon^2$

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

$\delta \leq \text{min}\{4\epsilon + \epsilon^2,4\epsilon - \epsilon^2\} =4\epsilon - \epsilon^2$

$ \epsilon \ngeq 4 $

but the epsilon-delta definition require there to be a $\delta$ for all $\epsilon$

so: how can you prove $\forall \epsilon$

Answer 1

Let $\epsilon\gt0$. Choose $\delta=2\epsilon$. Then \begin{align} 0\lt|x-4|\lt\delta\implies|\sqrt{x}-2|&=\frac{|x-4|}{\sqrt{x}+2}\\ &\lt\frac{\delta}{\sqrt{x}+2}\\ &\le\frac{\delta}2\\ &=\epsilon\\ \end{align} Hence $\lim_{x\to4}\sqrt{x}=2$ by definition.

Answer 2

The epsilon-delta method for calculating limits states that : $$\lim_{x\to a}f(x)=L$$ if $\forall \epsilon>0; \exists \delta(\epsilon)$ such that $$|x-a|<\delta(\epsilon) \implies |f(x)-L|<\epsilon$$

Imposing our conditions, we wish to show that there is a $\delta(\epsilon)$ which satisfies $$|x-4|<\delta(\epsilon)\implies|\sqrt x -2|<\epsilon$$

Use that $x-4=(\sqrt x -2)(\sqrt x+2)$ to help with this