# Existence of limits!

• Feb 2nd 2010, 11:30 AM
CrazyCat87
Existence 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| )
• Feb 2nd 2010, 11:54 AM
HallsofIvy
Quote:

Originally Posted by CrazyCat87
How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )

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 $\epsilon$ 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 $\epsilon$.
• Feb 2nd 2010, 01:01 PM
Drexel28
Quote:

Originally Posted by CrazyCat87
How would you go about showing that the limit as x approaches 0 of (x + sgn(x)) does not exist? (sgn(x) = x/|x| )

A slightly less messy approach would be to assume that $\lim_{x\to0}\left\{x+\text{sgn}(x)\right\}$ exists and equals $L$. Then, $L=L-0=\lim_{x\to0}\left\{x+\text{sgn}(x)\right\}-\lim_{x\to 0}x=\lim_{x\to0}\left\{x+\text{sgn}(x)-x\right\}=\lim_{x\to0}\text{sgn}(x)$. The combination of the limits was justified under the assumption that our limit exists. But the above then implies that $\lim_{x\to0}\text{sgn}(x)$ exists, which by following HallsOfIvy's example is just ridiculous.