Here is one way you could go:
1. Forget about epsilon to begin with!
2. Instead, try to bound the function expression, i.e, |\sqrt{x}-2| in terms of a delta!
3. Now, we can start with:
|\sqrt{x}-2}|=\frac{|\sqrt{x}-2|*|\sqrt{x}+2|}{|\sqrt{x}+2|}=\frac{|x-4|}{\sqrt{x}+2}
4. Now, consider the fraction we arrived at in 3, which is equal to our original function expression.
Can we BOUND this in terms of delta, if we ASSUME that we have the general bound 0<|x-4|<\delta?
Sure we can!
Evidently, we have: \frac{|x-4|}{\sqrt{x}+2}<\frac{\delta}{\sqrt{x}+2}
Agreed thus far?
5. WHAT ABOUT THE DENOMINATOR?
Remember that \sqrt{x}>0, so clearly, we have \sqrt{x}+2>2
But, therefore, we must have:
\frac{\delta}{\sqrt{x}+2}<\frac{\delta}{2}!
6. RECAPITULATION:
We have proven that IF 0<|x-4|<\delta, THEN we have:
|\sqrt{x}-2|<\frac{\delta}{2}
7. INTRODUCTION OF EPSILON:
Now, obviously, as long as delta fulfils \delta<2\epsilon, we get:
|\sqrt{x}-2|<\frac{\delta}{2}<\frac{2\epsilon}{2}=\epsilon
Thus, we have proved the continuity of our function at x=4, AND also, that by setting \delta<2\epsilon, we are ensured that our function expression will be bounded by epsilon.
8. TRANSLATION INTO CASE EPSILON=0.5
Just set delta=2*0.5=1 (yes, the equality works, too here), that is, we are in the interval 3<x<5.
Then, we are ensured that our expression is bounded by 0.5
