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.