In my assignment I have to prove the following limit, by definition:
$$\lim _{x \to 2}\sqrt{3x-2}=2 $$
I have made some calculations trying to prove it, and I've made some way but I'm afraid I'm wrong somewhere. Your help is needed, please.
$$|\sqrt{3x-2}-2|$$ $$=\frac{|(\sqrt{3x-2}|-2)*|(\sqrt{3x-2}+2)}{\sqrt{3x-2}+2}$$ $$=\frac{|3x-6|}{\sqrt{3x-2}+2}$$ $$=\frac{3|x-2|}{\sqrt{3x-2}+2}<\frac{3|x-2|}{\frac{1}{3}}=9*|x-2|$$
Now we choose $$\delta=\frac{\epsilon}{9}$$
And say that $$9*|x-2|<\delta$$ $$|x-2|<\frac{\epsilon}{9}<\epsilon$$
And we proved what's needed.
Is my solution correct?