Pick epsilon = 1/2, and see where that takes you.
2) You should know the limit is 3. Let \(\displaystyle \epsilon > 0\) You want to find \(\displaystyle \delta(\epsilon)>0\) ("a real number delta which may depend on epsilon") such that if \(\displaystyle 0<|x-1|<\delta\) then \(\displaystyle |\sqrt{x+8}-3| < \epsilon\).
Hint:
Play with: \(\displaystyle |\sqrt{x+8}-3| < \epsilon\). Attempt to get \(\displaystyle x-1\) by itself, trapped between two inequalities consisting of epsilony stuff, here given as triangle and square: \(\displaystyle \triangle < x-1 < \square\). Then let \(\displaystyle \delta=\min\{|\triangle|,|\square|\}\)
