Evaluate lim_o(|2x-1|-|2x+1|)/x

thax (Clapping)

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- Nov 11th 2009, 07:55 PMsakura hulimit question
Evaluate lim_o(|2x-1|-|2x+1|)/x

thax (Clapping) - Nov 11th 2009, 08:49 PMDrexel28
- Nov 11th 2009, 09:04 PMTKHunny
Kind of a surprising result, I think.

Abolute values can be tricky. It always helps if you can simply get rid of them. One way to get rid of them is to know when they are positive or negative.

In this case, there is a little leap of faith. It goes like this, as you approach zero, eventually, you will be rather close to zero. More to the point, eventually you will be closer than 1/2. 1/4, 1/8, 1/16, etc. or -1/4, -1/8, -1/16. This is VERY important, because IF we promise not to wander off by more than 1/2, we have the following delightful results.

For |x| < 1/2, 2x + 1 > 0 and |2x + 1| = (2x + 1)

For |x| < 1/2, 2x - 1 < 0 and |2x - 1| = (1 - 2x)

What can you do to the numerator with those two results?

Can you finish?