Uniform continuity and limits of sequences

Question

Bit stuck on this one so would appreciate any help.

Let $f : (0, 1)\to\mathbb R$ be uniformly continuous and $(x_n)$ and $(y_n)$ be sequences in $(0, 1)$ where $$\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = 0.$$

Show that $$\lim_{n\to\infty} f(x_n) − f(y_n) = 0.$$

Think I will have to use the triangle inequality but struggling to see where.

Answer 1

$f$ is uniformly continuous implies that for every $c>0$ there exists $d>0$ such that $|x-y|<c$ implies that $|f(x)-f(y)|<d$, there exists $N$ such that $n>N$ implies that $|x_n|<c/2$ and $|y_n|<c/2$, we deduce that $|x_n-y_n|<|x_n|+|y_n|<c$ it implies that $|f(x_n)-f(y_n)|<d$.

Answer 2

$|f(x)-f(y)|<\epsilon$, $x,y\in(0,1)$, $|x-y|<\delta$.

Now we know that $\lim_{n}(x_{n}-y_{n})=0$, so for some $N$, $|x_{n}-y_{n}|<\delta$, $n\geq N$.

Then $|f(x_{n})-f(y_{n})|<\epsilon$ for all such $n$.