Re: Evaluating the limit

On Mar 30, 2008 4:46 PM CT, conrad wrote:

My book evaluates the limit:
lim_{x->0} (|2x - 1| - |2x + 1|)/x

It offers the following reasoning:
For -1/2 < x < 1/2, we have 2x-1 < 0
and 2x + 1 > 0, so |2x - 1| = -(2x - 1)
and |2x + 1| = 2x + 1

lim_{x -> 0} [-(2x - 1) - (2x + 1)]/x = lim_{x->0} -4


What I fail to understand in their reasoning is
why consider |2x - 1| for values less than
zero but |2x + 1| for values greater than zero?

What your text is doing is examining the function |2x - 1|
and |2x + 1| on the interval (-1/2, 1/2). They choose
these points because the absolute value functions change
direction there. Moreover, you are evaluating the limit
within that interval.

I thought that both |2x - 1| and |2x + 1| should
have been considered for values less than 0
then |2x - 1| and |2x + 1| should have been
considered for values greater than zero.

Does not the interval (-1/2, 1/2) contain values less than
and greater than zero?

Maybe it would help you to explicitly write down the
functions. For example |2x - 1| = 0 when x = 1/2, so we
have

|2x - 1| = 2x - 1 if x > 1/2, and

|2x - 1| = -(2x - 1) if x < 1/2.

Thus, |2x - 1| = -(2x - 1) on the interval (-oo, 1/2).
Similarly, |2x + 1| = 0 when x = -1/2, so we have

|2x + 1| = 2x + 1 if x > -1/2, and

|2x + 1 = -(2x + 1) if x < -1/2.

It follows that |2x + 1| = 2x + 1 on the interval (-1/2,
oo). Putting all these facts together, we see that on
the interval (-1/2, 1/2) we have |2x - 1| = -(2x - 1) and
|2x + 1| = 2x + 1. Hence, your limit becomes

lim_{x -> 0} [-(2x - 1) - (2x + 1)] / x =

lim_{x -> 0} (-2x + 1 - 2x - 1) / x =

lim_{x -> 0} -4x / x = -4.

Regards,
Kyle Czarnecki

Regards,
Kyle Czarnecki
.

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