# Limits of functions; approaching from left and right

1. Aug 22, 2008

### Niles

1. The problem statement, all variables and given/known data
Hi all.

Please take a look at this function:

$$f(x) = \left\{ {\begin{array}{*{20}c} { - x - 1\,\,\,\,\,{\rm{for}}\,\,\,\,\, - 2 \le x < 0} \\ { - x + 1\,\,\,\,\,{\rm{for}}\,\,\,\,\,0 \le x < 2} \\ \end{array}} \right.$$

According to my book we have that:

$$\begin{array}{l} f(0 - ) = - 1 \\ f(0 + ) = 1 \\ \end{array}$$

My question is: Why does f(0-) = -1, when f(x) = -x+1 for x = 0?

EDIT: I did write a title, but when I editted my post, it must apparently have been resetted. The title was: "Limits of functions; approaching from left and right". Perhaps a moderator can insert the title? Done - cristo

Last edited by a moderator: Aug 22, 2008
2. Aug 22, 2008

### HallsofIvy

Staff Emeritus
Remember that $f(0^-)= \lim_{x\rightarrow 0^-} f(x)$ means the "limit from below". That is, we are looking at what f(x) is for x close to 0 but less than 0. For example, if x= -0.0000001, then x- 1= -0.0000001- 1= -1.0000001. That in itself does not tell us that the limit is -1 but it should be clear that no matter how close to 0 x is as long as it is not equal to 0!

The definition of "$\lim(x\rightarrow a}f(x)= L$" is "Given any $\epsilon> 0$ there exist $\delta> 0$ such that if $0< |x- a|< \delta$, then $|f(x)-L|< \epsilon$.

Do you see that "0< |x- a|" part? (Which, I confess, I like many teachers often forget to write!) One of the crucially important things you need to learn about limits is this:
The $\lim_{x\rightarrow a} f(x)$ has absolutely no dependence on f(a)!

If I define f(x)= 3x- 2 for x not[/b\] 0, then no matter what the value of f(a) is, or even if f(a) is undefined, $\lim_{x\rightarrow 0} f(x)= -2$.

One reason that is hard to understand is that we often learn to find limit by evaluating f(a)! But that only works for continuous functions. In fact, that is the definition of "continuous"! Because continuous functions are so nice, it happens that our methods of writing functions have grown up to make it easy to write continuous functions so we tend to automatically think that functions are continuous- that is NOT true. In a very specific sense, "almost all" functions are not continuous anywhere.

3. Aug 22, 2008

### Niles

It there was a "Thread solved"-button, I would have pressed it.

Thanks!