- #1

maupassant

- 10

- 0

Suppose one wants to prove that lim (3x²-x) = 10 as x approaches 2.

Taking the definition of a limit we should have:

"If 0 < |x-2| < d then it follows that |f(x) - 10| < e"

Proving this statement, we can say |f(x)-10| < e

iff |3x²-x-10| <e

iff |3x+5|.|x-2| <e

iff |x-2| < e/(|3X+5|)

The problem that arises now is: how can one make the "x" in the denominator disappear?

Because we need a delta in function of only an epsilon.

So we could say: set d<=1 then it follows

|x-2|<d<=1

|x-2|<1

-1<x-2<1

1<x<3

**We know now that our x should lie between 1 and 3. But now comes the thing I fail to understand well: Why does one select 3 which is larger than the x should be. Why could we not just select 2 or 2.5 or something like that. So why should we have**

|x-2| < e/(|3.3+5|)

|x-2| < e/(14)

|x-2| < e/(|3.3+5|)

|x-2| < e/(14)

Finally we can set d=min{1, e/(14)} and we should have a solution for

"If 0<|x-2|< d=e/(14) then |f(x) - 10|<e" and have proved that indeed

lim (3x²-x) = 10 as x approaches 2.

Could somebody tell me why I always should use that maximum border on x (i.c. 3) ? That's something that's been annoying me for about a week now!

Thank you very much!

M.