Do Right and Left Hand Limits Exist for All Increasing Functions?

  • Thread starter Thread starter SomeRandomGuy
  • Start date Start date
  • Tags Tags
    Analysis
SomeRandomGuy
Messages
55
Reaction score
0
Been bangin my head against the wall all weekend thinkin about this question:

Let f:R->R be an increasing function. Proce that lim x->c+f(x) and lim x-c-f(x) (right and left hand limits) must each exist at every point c in R.

There's more to the question, but if I can get this part solved, I'm sure the rest won't be trouble.

My original idea was to prove this by contradiction, assuming the limits don't exist, and showing this violates the increasing aspect. However, I've come to deadends each time. Proving it directly seems very difficult as well.
 
Physics news on Phys.org
Has your text defined left/right-handed limits of f at x in terms of what happens to f\left( x_n\right) for alll sequences that converge to x which are strictly less than/greater than x?
 
benorin said:
Has your text defined left/right-handed limits of f at x in terms of what happens to f\left( x_n\right) for alll sequences that converge to x which are strictly less than/greater than x?

Yes,

limx->c+ f(x) = L if limf(x_n)=L for all x_n > c where lim(x_n)=c.

Similar definition for for limx->c- f(x).

Since f was defined as an increasing function, then x_n is either increasing or decreasing depending on what limit we are taking. How can we conclude that limf(x_n) = L? Still kind of confused. I'll go work on it and check back in the morning.

Thanks for the help.
 
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...

Similar threads

Replies
2
Views
1K
Replies
13
Views
3K
Replies
7
Views
1K
Replies
6
Views
1K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
2
Views
1K
Back
Top