# 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.

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If you have results about continuous functions you're allowed to use, then by far the easiest is to note that $f(x)=\sqrt{9-x}+3$ is continuous in a neighborhood of $0$ (namely $(-\infty,9)$) and therefore $\lim_{x\to 0}f(x) = f(0)$. – Henning Makholm Oct 4 '12 at 16:15
Sorry I don't know that result, or anything about continuous functions. All I know is the definition of a $\epsilon-\delta$ definition of a limit. – tk2 Oct 4 '12 at 16:17

$|\sqrt{9-x}-3|=\frac{|x|}{\sqrt{9-x}+3}\leq\frac{1}{3}|x|$, you can set $\delta=3\epsilon$.
Firstly, observe that for any $x \in (- \infty,9]$, we have $$|\sqrt{9 - x} - 3||\sqrt{9 - x} + 3| = |(9 - x) - 9| = |x|.$$ 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 $\delta := \min(9,3 \epsilon)$.
Hint: Multiply by $\sqrt{9-x}-3$ by $\dfrac{a}{a}$ where $a=\sqrt{9-x}+3$.