How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )

Printable View

- February 2nd 2010, 12:30 PMCrazyCat87Existence of limits!
How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )

- February 2nd 2010, 12:54 PMHallsofIvy
If x< 0, then x+ sgn(x)= x- 1. If x> 0, then x+ sgn(x)= x+1. For x very close to 0 and negative, x+ sgn(x) is very close to -1. For x very close to 0 and positive, x+ sgn(x) is very close to 1. For any putative limit, a, take to be the smaller of |a-1|/2 and |a+1|/2. Depending on which you use, there will be either positive x or negative, arbitrarily close to x such that |x+ sign(x)- a| is larger than .

- February 2nd 2010, 02:01 PMDrexel28
A slightly less messy approach would be to assume that exists and equals . Then, . The combination of the limits was justified under the assumption that our limit exists. But the above then implies that exists, which by following

**HallsOfIvy**'s example is just ridiculous.