# Thread: limits and uniformly continuous

1. ## limits and uniformly continuous

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|
< epsilon.

Thus, the right hand limit of f exist

Does this make sense?

2. Originally Posted by inthequestofproofs
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|
< epsilon.

Thus, the right hand limit of f exist

Does this make sense?
Not really. First, we don't know what your $\displaystyle L$ is; second, $\displaystyle |f(x)-L|+|L-f(y)|<|f(x)-L|$ is more than dubious.

Since you have to prove the existence of a limit and your function is not monotone, you should use Cauchy condition. This should be pretty straightforward (no computation at all).

3. sorry for so many questions, but the cauchy condition is for sequences. I don't know how to apply it for a function. Could you explain?

4. Originally Posted by inthequestofproofs
sorry for so many questions, but the cauchy condition is for sequences. I don't know how to apply it for a function. Could you explain?
You have the same criterion as for sequences.

Fact (Cauchy property for functions) : $\displaystyle f(t)$ has a limit when $\displaystyle t\to a^+$ if, and only if, for every $\displaystyle \epsilon>0$, there is $\displaystyle t_0>a$ such that, for any $\displaystyle t,t'\in(a,t_0)$, $\displaystyle |f(t)-f(t')|<\epsilon$.

Proof:
a) Assume $\displaystyle f(t)\to_{t\to a^+}\ell$. Let $\displaystyle \epsilon>0$. There is $\displaystyle t_0>a$ such that, if $\displaystyle a<t<t_0$, $\displaystyle |f(t)-\ell|<\epsilon/2$. Then, for all $\displaystyle a<t,t'<t_0$, $\displaystyle |f(t)-f(t')|\leq |f(t)-\ell|+|\ell-f(t')|\leq \epsilon$.

b) Assume Cauchy condition : for every $\displaystyle \epsilon>0$, there is $\displaystyle t_0>a$ such that, for any $\displaystyle a<t,t'<t_0$, $\displaystyle |f(t)-f(t')|<\epsilon$. Choose a particular sequence $\displaystyle (u_n)_{n\in\mathbb{N}}$ such that $\displaystyle u_n\to_n a^+$. Then the assumption shows that $\displaystyle (f(u_n))_{n\in\mathbb{N}}$ is a Cauchy sequence (I let you check that). Therefore, it has a limit $\displaystyle \ell\in\mathbb{R}$ : $\displaystyle f(u_n)\to_n \ell$.
Let us show that $\displaystyle f(t)\to_{t\to a^+} \ell$. Let $\displaystyle \epsilon>0$. Let $\displaystyle t_0$ be given by the assumption. There is $\displaystyle n_0$ such that $\displaystyle n\geq n_0$ implies $\displaystyle a<u_n<t_0$ (because $\displaystyle u_n\to_n a^+$) and $\displaystyle |f(u_n)-\ell|<\epsilon$ (because $\displaystyle f(u_n)\to_n\ell$). For every $\displaystyle a<t<t_0$, we have $\displaystyle |f(t)-\ell|\leq |f(t)-f(u_n)|+|f(u_n)-\ell|\leq 2\epsilon$ (first term because of Cauchy condition since $\displaystyle a<u_n,t<t_0$). This concludes.

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In your problem, the uniform continuity immediately gives the Cauchy condition.