I need to show that if f is uniformly continuous on (a,b], then the right hand limit of f(x) exists.
This is what i have so far:
If f is uniformly continuous on (a,b], given epsilon >0, there is a delta > 0 s.t. |x-y| < delta -> |f(x)-f(y)| < epsilon for all x,y in (a,b]
Let x be in (a, a + delta).
|f(x) - f(y)|
<= |f(x) - L|+|L-f(y)|
< |f(x) - L|
Thus, the right hand limit of f exist
Does this make sense?
Fact (Cauchy property for functions) : has a limit when if, and only if, for every , there is such that, for any , .
a) Assume . Let . There is such that, if , . Then, for all , .
b) Assume Cauchy condition : for every , there is such that, for any , . Choose a particular sequence such that . Then the assumption shows that is a Cauchy sequence (I let you check that). Therefore, it has a limit : .
Let us show that . Let . Let be given by the assumption. There is such that implies (because ) and (because ). For every , we have (first term because of Cauchy condition since ). This concludes.
In your problem, the uniform continuity immediately gives the Cauchy condition.