Hey,

I wanted to ask how can I proof the following statement using epsilon delta.

lim x/[x] = 1

x->3+

*[x] means the whole part of x, for example : [3.5] = 3, [-3.5] = -4, [3] = 3.

I do know how to proof normal statements using epsilon delta and so far I got to this point where I am stuck:

|x/[x] -1| < e

...

|x| < e/|1/[x] - 1/x|

After that i decided to limit the neighborhood of 3 to a radius of 1, therefore:

3 < x < 4

...

0<1/[x] - 1/x < 1/12

therefore:

|x| < 12e

Now I only have to choose an L so that if |x| < L, than |x| < 12e:

3 < x < 3 + L

But I cant find a proper L so that |x|< 12e.

I thought using L = 12e - 3, and than :

3 < x < 3 -3 +12e

|x| < 12e

But L must be larger than 0 for each e, so L = 12e -3 is no good.

Thanks a lot,

Michael.