Limits of functions; approaching from left and right

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The discussion focuses on understanding limits of a piecewise function at a specific point, x = 0. The function is defined differently for intervals around zero, leading to different left-hand and right-hand limits: f(0-) = -1 and f(0+) = 1. The key point is that the limit as x approaches a value does not depend on the function's value at that point, emphasizing that limits are concerned with values approaching the point from either side. This distinction is crucial for grasping the concept of limits, especially for functions that are not continuous. Understanding this helps clarify why f(0-) results in -1 despite the function's definition at x = 0.
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Homework Statement


Hi all.

Please take a look at this function:

<br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> { - x - 1\,\,\,\,\,{\rm{for}}\,\,\,\,\, - 2 \le x &lt; 0} \\<br /> { - x + 1\,\,\,\,\,{\rm{for}}\,\,\,\,\,0 \le x &lt; 2} \\<br /> \end{array}} \right.<br />

According to my book we have that:

<br /> \begin{array}{l}<br /> f(0 - ) = - 1 \\ <br /> f(0 + ) = 1 \\ <br /> \end{array}<br />

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
 
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Niles said:

Homework Statement


Hi all.

Please take a look at this function:

<br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> { - x - 1\,\,\,\,\,{\rm{for}}\,\,\,\,\, - 2 \le x &lt; 0} \\<br /> { - x + 1\,\,\,\,\,{\rm{for}}\,\,\,\,\,0 \le x &lt; 2} \\<br /> \end{array}} \right.<br />

According to my book we have that:

<br /> \begin{array}{l}<br /> f(0 - ) = - 1 \\ <br /> f(0 + ) = 1 \\ <br /> \end{array}<br />

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

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&gt; 0 there exist \delta&gt; 0 such that if 0&lt; |x- a|&lt; \delta, then |f(x)-L|&lt; \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.
 
It there was a "Thread solved"-button, I would have pressed it.

Thanks!
 
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