Well, how then did you find it?But my problem is that how do I verify/show that my delta works?
In order to prove that , you need to show that for any there exists a such that for all x with the following inequality holds:
Now, simplify the left side to get
Pull out a positive factor of 3 like this
and divide by 3 to get
which is equivalent to
Now all these inequalities are equivalent. So if the last inequality is satisfied, which is the case if you set and require that , then the first of these inequalities, the one that you have to show, also holds.
Now, here again, you start with the inequalityAnd there is another question where I am not sure how to approach it after simplifying it
lim x->2 (x^3)=8
and try to figure out how requiring , for a suitably chosen , guarantees that this inequality holds.
You have already shown a useful first step toward that goal, namely, factoring out x-2
Now the problem that we face is to show that for |x-2| sufficiently small, the second factor on the left cannot get arbitrarily large. In other words, we need a way to bound that second, quadratic factor on the left with a constant, M, say.
How could we do that?
Well, by completion of the square. Thus the quadratic polynomial and its absolute value are the same (because the polynomial is always positive) and it assumes its smallest value 3 at x=-1.
So lets therefore already limit x to satisfy |x-2|<1, i.e. x must be from the interval (1,3). For such values of x, the quadratic factor assumes its largest value at x=3, which gives us the required upper bound .
Finally, divide the above inequality by M on both sides to get your .
(We must make sure that , because otherwies our choice of M as an upper bound for the quadratic factor on the left side of the inequality would not be valid.)