If $f$ is a continuous function, then $f$ is symmetrically continuous.

Question

I am trying to prove the following:

If $f$ is a continuous function, then $f$ is symmetrically continuous.

My issue is interpreting the definition of symmetrically continous. It states that a function is symmetrically continuous if:

$\lim_{h\to 0^+}{[f(x+h)-f(x-h)]}=0.$

I am not sure how to turn this into an epsilon delta definiton. My attempt:

A function $f$ is symmetrically continuous if: for all $\epsilon >0$ there exists $\delta>0$ such that $|f(x+h)-f(x-h)|<\epsilon$ whenver $0<h<\delta$.

Answer 1

Take $\varepsilon>0$. Then there is a $\delta>0$ such that$$|y-x|<\delta\implies\bigl|f(y)-f(x)\bigr|<\frac\varepsilon2.$$Therefore, if $0<h<\delta$,\begin{align}\bigl|f(x+h)-f(x-h)\bigr|&\leqslant\bigl|f(x+h)-f(x)\bigr|+\bigl|f(x)-f(x-h)\bigr|\\&<\frac\varepsilon2+\frac\varepsilon2\\&=\varepsilon.\end{align}