The function $$f(x) = \left\{
\begin{array}{l l}
x-2n, & \quad \text{if } {2n \leq{x} \leq{2n+1}}, \;\; {n \in \mathbb{Z}}\\
2n-x, & \quad \text{if } {2n-1 \leq{x} \leq{2n}}, \;\; {n \in \mathbb{Z}}
\end{array} \right.$$
is not differentiable only at points $x_n=n, \;\; {n \in \mathbb{Z}},$ but everywhere continuous. You are right that $\lim\limits_{x \to 2n^{-}} f(x) = 0 = \lim\limits_{x \to 2n^{+}} f(x).$ In addition,$\lim\limits_{x \to (2n+1)^{-}} f(x) = 1 = \lim\limits_{x \to (2n+1)^{+}} f(x)$ and $f(x)$ is continuous on every open interval $(2n,\; 2n+1).$
Fact 1. A uniformly continuous function on $(0,\infty)$ is Lipschitz for large distances, which can be expressed concisely by the inequality
$$|f(x)-f(y)|\le L|x-y|+M,\quad x,y>0 \tag{1}$$
where $L$ and $M$ are constants. This is a general and useful fact, so it's worth recording regardless of this problem. Proof: there is $\delta>0$ such that $|f(x)-f(y)| \le 1$ whenever $|x-y|\le \delta$. Divide $[x,y]$ into intervals of size at most $\delta$; we need no more than $1+|y-x|/\delta$ of these. Then $|f(x)-f(y)|\le 1+|y-x|/\delta$, which gives (1). $\quad \Box$
In particular, (1) implies that $(f(x)-f(1))/(x-1)$ remains bounded as $x\to \infty$. This excludes $\alpha>1$ from consideration.
Furthermore, consider the sequence $x_n=(\pi/2+\pi n)^{1/\beta}$. Observe that
$|x_{n+1}-x_n| \le C n^{\frac{1}{\beta}-1}$ and $|f(x_n)-f(x_{n+1})|\ge c n^{\alpha}$ where $C$ and $c$ are positive constants independent of $n$. By (1), $\alpha\le \frac{1}{\beta}-1$.
Summarize the necessary conditions obtained so far:
$$0<\alpha\le 1, \qquad 0<\beta \le \frac{1}{\alpha+1}\tag{2}$$
Here's an interesting fact: under conditions (2), the second derivative of $f$ is bounded at infinity. Indeed, the largest term in $f''$ has $x^{\alpha+2\beta-2}$, and
$$\alpha+2\beta-2 \le \frac{\alpha^2-\alpha}{\alpha+1}\le 0$$
Fact 2. If $f$ is uniformly continuous on $(0,\infty)$ and $f''$ is bounded at infinity, then $f'$ is also bounded at infinity.
Proof: there is $\delta>0$ such that $|f(x)-f(y)| \le 1$ whenever $|x-y|\le \delta$. By the mean value theorem, this implies $|f'|\le 1/\delta$ at some point within the interval $[x,x+\delta]$. But then $|f'(x)| \le 1/\delta+ \delta \sup_{[x,x+\delta]} |f''|$, which is a uniform bound on $f'$. $\quad \Box$
It is not hard to see that $f'$ is bounded at infinity if and only if
$$ \alpha +\beta \le 1 \tag{3} $$
So, (3) is necessary for uniform continuity. It is also sufficient, since bounded derivative implies Lipschitz implies uniformly continuous on $[1,\infty)$; the interval $[0,1]$ is compact, so it's not a problem.
Concerning equicontinuity: on $[1,\infty)$, the family described by (3) is equicontinuous because the derivative $f'$ is uniformly bounded. There is an issue at $0$, where $f$ behaves like $x^{\alpha+\beta}$. You don't want to allow this exponent to be arbitrarily small. So,
$$ \epsilon \le \alpha +\beta \le 1 \tag{4} $$
gives an equicontinuous family, for every $\epsilon>0$.
Best Answer
Set $g(t)=f(t,t)$ and look at the point $(1/3,1/3)$, the point is on the curve $y=3x^{2}$ and if $t\rightarrow 1/3^{-}$, then $f(t,t)\rightarrow 0$ and if $t\rightarrow 1/3^{+}$, then $f(t,t)\rightarrow 1$, $g$ is not continuous at $t=1/3$.