<s>[INST] Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt: 
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \frac{1}{|\sqrt{x+1}+2|}$
Here, the second term is the nuisance so maybe do something like the following:
Suppose we want $|x-3| < 1$. Then, $2 < x < 4$. 
So $|x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3|$
Looks like we want $\delta = \min((\sqrt{3}+2)\epsilon, 1)$
Then $|\sqrt{x+1} - 2| = |x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3| < \frac{1}{\sqrt{3}+2} \cdot (\sqrt{3} + 2)\epsilon = \epsilon$
 [/INST] Your derivation is correct (I believe, it looks right but I didn't check every detail), but you are going for too much. To prove the limit statement, you don't need to identify specifically the largest $\delta$ that works for each $\epsilon$. You just need to prove there is some positive $\delta$ that will work. And you only need to prove it for "small" $\epsilon$ (it automatically follows for larger values). So, assuming $\epsilon < 1$, $\delta = \epsilon$ would work, because the nuisance second term is less than 1.
</s>