Proving f(x) has a limit at all points not odd

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1.

Define f:R→R as follows:
f(x)= x- ⌊x⌋ if ⌊x⌋ is even.
f(x)= x- ⌊x+1⌋ if ⌊x⌋ is odd.
Determine those points where f has a limit and justify your conclusions (using δ and ε).



Homework Equations





3.

The attempt at a solution involved a graph of the situations of f(x). With this graph my group and I determined f(x) has a limit at x_0 iff x_0 is not an odd integer. However, we hare having a hard time proving it, and, although the graph says we are right, we must use δ and ε in a formal proof.

 
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angelpsymon said:
1.

Define f:R→R as follows:
f(x)= x- ⌊x⌋ if ⌊x⌋ is even.
f(x)= x- ⌊x+1⌋ if ⌊x⌋ is odd.
Determine those points where f has a limit and justify your conclusions (using δ and ε).



Homework Equations





3.

The attempt at a solution involved a graph of the situations of f(x). With this graph my group and I determined f(x) has a limit at x_0 iff x_0 is not an odd integer. However, we hare having a hard time proving it, and, although the graph says we are right, we must use δ and ε in a formal proof.

You have used some special symbols that do not show on my internet reader. What are they (describe in words)? Is it the "floor" symbol: \floor{x}
 
yes, it is the floor symbol, the others are delta and epsilon
 
angelpsymon said:
1.

Define f:R→R as follows:
f(x)= x- ⌊x⌋ if ⌊x⌋ is even.
f(x)= x- ⌊x+1⌋ if ⌊x⌋ is odd.
Determine those points where f has a limit and justify your conclusions (using δ and ε).



Homework Equations





3.

The attempt at a solution involved a graph of the situations of f(x). With this graph my group and I determined f(x) has a limit at x_0 iff x_0 is not an odd integer. However, we hare having a hard time proving it, and, although the graph says we are right, we must use δ and ε in a formal proof.

You have used special symbols that do not show on my internet reader. Is that the "floor function", f(x)= largest integer less than or equal to x? If so then the fact that floor(x+1)= floor(x)+ 1 should be helpful.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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