Is it reasonable to say that the following limit exists? (posting below in a minute.)

Question

mendicant_bias · Answer

Let's say that we have a piecewise defined function that looks like this:

$$y = x + 2,||| (-\infty, -4]$$$$y =-2,||| [-4, \infty)$$
So the function meets like this:

mendicant_bias · Answer

|dw:1373426904470:dw|

mendicant_bias · Answer

While it's typically said in textbooks that despite things like removable discontiunities or "sharp" breaks, e.g. the graph of y = |x|, the limits for the discontinuities discussed are said to exist because despite the value at the point in question being undefined, both sides of the function "approach" the same value; both left-handed and right-handed limits approach the same value. In this case, while the left-handed limit approaches the same value of the right-handed limit in a sense, the right-handed limit that's on the constant function doesn't "approach" anything; it's just a horizontal line. So if I took the limit at that point, e.g. 

$$\lim_{x \rightarrow -2}(both functions)$$
Would that limit exist?

hero · Answer

The limit exists. The only way it wouldn't exist is if each "piece" approached different values.

hero · Answer

at x = -2

mendicant_bias · Answer

Okay, Just didn't want to assume but I thought that was the case; both my teacher and my textbook kind of heavily laid down the whole "both approach" deal, and in this case it can be argued that the constant function doesn't "approach" anything. And why doesn't it have a true sharp corner/what makes a true sharp corner?

mendicant_bias · Answer

I thought that there was a limit on the sharp corner. I checked wolframalpha (lim |x| as x approaches 0), and it gave me the limit as being zero.

mendicant_bias · Answer

http://www.wolframalpha.com/input/?i=lim+ |x|+as+x+approaches+0

mendicant_bias · Answer

Damn, guess the full thing won't turn into a hyperlink.

hero · Answer

Hmm. Well, as I said, as long as both sides are approaching the same value, then the limit exists.

hero · Answer

I think for sharp corners, the rule is, it is not differentiable at the point.

mendicant_bias · Answer

Agreed on that, just wasn't sure regarding limits; I know that a derivative is a special type of limit and still thought that limits could be taken on points where it's not differentiable. Don't know much about the uses of limits other than derivatives, though.