I dont get the last step of the proof that explains that
Let $f:(a,b)\rightarrow \mathbb{R}\ $is continuous. $f\ $uniformly continuous if $\ \lim _{x\to a^+}\left(f\left(x\right)\right)\ $ and $\ \lim _{x\to b^-}\left(f\left(x\right)\right)\ $exist and are finite.
The proof went as follows;
Let $f\ $ uniforlmy continuous
$xn\ $sequence in $(a,b)\ $ with $xn \rightarrow a$ and set $l=\lim _n\left(f\left(xn\right)\right)$
$yn\ $sequence in $(a,b)\ $ with $yn \rightarrow a$ and set $l'=\lim _n\left(f\left(xn\right)\right)$
Since $f\ $uniformly continuous for $xn-yn \rightarrow 0\ $ we have $f(xn)-f(yn)\rightarrow0\ $
There for $l=l'$
So for every $yn\ $that belongs to $(a,b)\ $with $yn \rightarrow a$ we have $f(yn)\rightarrow l$
Finaly $\ \lim _{x\to a^+}\left(f\left(x\right)\right)=l $
i dont get a couple of things about it. The first one is i dont understand the last step where we conclude that the limit exists and its l. Secondly i dont get why we need to get in the process of creating a yn in the first place, since the conclusion is exactly the same as we stated for xn. Ive benn looking at it for a while and i cant make sense of it please help
If $x_n \to a$ then $f(x_n)$ is easily seen to be a Cauchy sequence so it is convergent, say to $l$. To show that $\lim_{x \to a} f(x)$ exists we have to make sure that we don't get different limits for different sequences $(x_n)$ converging to $a$. So we take another sequence $y_n$ tending to $a$ and show that the limit is the same. If $l'$ is the limit of $f(y_n)$ then, using uniform continuity again we see that $f(x_n)-f(y_n) \to 0$ which gives $l=l'$.
For the last step assume that $f(x)$ does not tend to $l$ as $ x\to a$. Then there exists $\epsilon >0$ such that whatever $\delta >0 $ we take there exists $x$ with $a <x<a+\delta$ but $|f(x)-l| \geq \epsilon$. Taking $\delta =\frac 1n$ we construct $x_n$ with $a <x_n<a+\frac 1 n $ but $|f(x)-l| \geq \epsilon$. But this contradicts the fact that $f(x_n) \to l$.