Is this delta epsilon proof ok? (single variable)

Question

I have to prove that $$\lim_{x\to0}\; \sqrt {x+3}=\sqrt 3$$

$$ \forall \; \epsilon >0 \;\;\; \exists \; \delta>0:\;0<|x|<\delta \;\rightarrow \; |\sqrt {x+3}-\sqrt {3}|<\epsilon$$ So I started getting rid of the radical in the numerator \begin{align*} |\sqrt {x+3}-\sqrt {3}|&=\dfrac{|\sqrt {x+3}-\sqrt {3}|\cdot |\sqrt {x+3}+\sqrt {3}|}{|\sqrt {x+3}+\sqrt {3}|} \\ \\ &=\dfrac{|(\sqrt {x+3}-\sqrt {3})\cdot (\sqrt {x+3}+\sqrt {3})|}{|\sqrt {x+3}+\sqrt {3}|} \\ \\ &=\dfrac{|x+3-3|}{|\sqrt {x+3}+\sqrt {3}|} \\ \\ &=\dfrac{|x|}{|\sqrt {x+3}+\sqrt {3}|} \\ \\ &\leq |x| \quad (\text{I use the fact that the denominator is positive}) \\ \\ &< \delta \end{align*} $ \therefore \delta = \epsilon $

Is it ok?

Answer 1

The conclusion $\delta = \epsilon$ is not okay. You are not writing a proof of $\delta=\epsilon$.

And the last step should be $< \epsilon$ because you want to show that $|\sqrt{x+3}-\sqrt{x}| < \epsilon$.

What you mean with writing "$\therefore \delta = \epsilon$" probably is that we can take $\delta=\epsilon$. But then write it in the beginning of the proof.

A recommended structure of a limit proof:

Given $\epsilon>0$, let $\delta = \cdots$. Then, whenever $|x-a|<\delta$ we have $$|f(x)-a| = \cdots < \epsilon.$$