Saying "there is NO such value for

" is saying that the limit does not exist- and it obviously does!

If |f(x)- 6|< 2, then -2< f(x)- 6< 2 so 4< f(x)< 8. Drawing horizontal lines on the graph at y= 4 and y= 8, I see that the line y= 4 crosses the graph at (2, 4) and the line y= 8 crosses the graph at (8, 8). If I were to draw vertical lines at x= 2 and x= 8, I would form a rectangle with the graph completely inside the rectangle. That shows that 4< f(x)< 8, which is the same as |f(x)- 6|< 2, for 2< x< 8. Subtracting 4 from each part, -2< x- 4< 4. Since |-2|< |4|, that will be true as long as |x- 4|< |-2|= 2. We can take

.

For (b), do the same thing. |f(x)- 6|< 1 is the same as -1< f(x)- 6< 1 and then 5< f(x)< 7. Drawing horizontal lines on the graph at y= 5 and y= 7, I see that the graph crosses y= 5 at (3, 5) and y= 7 at (6, 7). That is, 5< f(x)< 7, which is the same as |f(x)- 6|< 1, as long as 3< x< 6 or -1< x- 4< 2. Since |-1|< 2, that will be true as long as |x- 4|< 1. We can take

.