<s>[INST] Prove that $\lim_{x \rightarrow 0} \sqrt{9-x} + 3 = 6$ Prove that $\lim_{x \rightarrow 0} \sqrt{9-x} + 3 = 6$
So far my difficulty is trying to find a $\delta$ that will allow for this function to be less than $\epsilon$.  I keep getting that $|\sqrt{9-x}+3-6|=|\sqrt{9-x}-3| < \epsilon$, but from here I do not know how to deal with finding  $\delta$.  Thanks in advance and any help is appreciated.
 [/INST] Firstly, observe that for any $ x \in (- \infty,9] $, we have
\begin{equation}
|\sqrt{9 - x} - 3||\sqrt{9 - x} + 3| = |(9 - x) - 9| = |x|.
\end{equation}
Fix an $ \epsilon > 0 $. Then choosing any $ x $ that satisfies $ |x| < \min(9,3 \epsilon) $, we obtain
\begin{align}
|\sqrt{9 - x} - 3| &= \frac{|x|}{\sqrt{9 - x} + 3} \\
                   &= |x| \cdot \frac{1}{\sqrt{9 - x} + 3} \\
                   &< 3 \epsilon \cdot \frac{1}{3} \\
                   &= \epsilon.
\end{align}
Hence, we can set $ \delta := \min(9,3 \epsilon) $.
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