Uniform continuity of function Let $f:( 0, ∞)→R$ such that 
$f(x) = 1/x $  then we know it is not uniformly continuous. As shown in graph, for any given $ε >0$, the same $δ$ does not work everywhere on graph. That is, $δ$ changes for same $ε$ for the different portion of graph, So that function is not uniformly continuous.
Now, if we consider $g(x) = 1/x $ in $(a, ∞)$ where $a > 0$ then too, I have different $δ$, for same $ε$ in different portion of graph. So why $g(x) = 1/x$ is  uniformly continuous on $(a,∞)$ where $a> 0$? 
 A: If we consider the interval $I=(a,\infty)$, then we see that, for any $x,x' \in I$:
\begin{align*}
\lvert f(x) - f(x') \rvert &= \left\lvert \frac{1}{x} - \frac{1}{x'} \right\rvert \\
&= \left\lvert \frac{x'-x}{x\cdot x'} \right\rvert \\
&< \frac{1}{a^2} \cdot \left\lvert x'-x \right\rvert \\
\end{align*}
For any $\epsilon > 0$, we can choose $\delta = a^2\epsilon$ and see that:
$$ \lvert x - x' \rvert < \delta \implies \lvert f(x) - f(x') \rvert < \epsilon \qquad \forall x,x' \in I$$
By definition, $f(x)$ is uniformly continuous on $I$.
To answer your question about why concerning the $\delta$ - the $\delta$ is not unique. Choosing a smaller $\delta$ works. Uniform continuity means that we can choose a $\delta$ that works regardless of which points are chosen - dependent only on $\epsilon$.
A: Let $\epsilon > 0$ and $g : (a,\infty) \to \mathbb R$ given by $g(x) = 1/x$. Take $\delta = a^2\epsilon$. Note that for any $x,y \in (a,\infty)$, $xy \ge a^2$ so $1/(xy) \le 1/a^2$. Then if $|x-y| < \delta$,
$$
\left| \frac{1}{x} - \frac{1}{y} \right| = \frac{|y-x|}{xy} \le \frac{|y-x|}{a^2} < \frac{\delta}{a^2} = \epsilon
$$
A: Suppose it is not uniformly cont on $(a,\infty)$. Then there exist sequence $u_n$ and $v_n$ such that $|u_n-v_n|\to 0$ but $|f(u_n)-f(v_n)|\not\to 0$. However we have $|f(u_n)-f(v_n)| =|1/u_n -1/v_n| =|\dfrac{v_n-u_n}{u_nv_n}|$. As long as $u_nv_n$ doesn't go to zero(because of our interval), we can evaluate the limit of the top and bottom separately. The numerator limits to zero, and the denominator will be bounded by $1/a$ and $0$. Thus $|f(u_n)-f(v_n)|\to 0$. This is a contradiction, thus $f$ is uniformly continuous on the interval.
Intuitively, uniform continuity means that the function cannot get too steep. As slope increases, the ratio of change in y to change in x increase. In epsilon-delta terminology, this means that our function cant have an infinitely large epsilon for a infinitely small delta. In this case the "steep" part of the function happens in the limit towards zero. When we choose the interval to $(0,\infty)$ we lose uniform continuity. But when we cap the interval at $a$, the slope of our function can be bounded, and so it is uniformly continuous.
