<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] Let $x = y+3$,
so that
$x \to 3$
is the same as
$y \to 0$.
$\begin{array}\\
\sqrt{x+1}-2
&=\sqrt{y+4}-2\\
&=(\sqrt{y+4}-2)\dfrac{\sqrt{y+4}+2}{\sqrt{y+4}+2}\\
&=\dfrac{(\sqrt{y+4}-2)(\sqrt{y+4}+2)}{\sqrt{y+4}+2}\\
&=\dfrac{(y+4)-4}{\sqrt{y+4}+2}\\
&=\dfrac{y}{\sqrt{y+4}+2}\\
&\to 0 \text{ as } y \to 0\\
\end{array}
$
Move on - nothing original here.
</s>