Choosing the right $\delta$ for uniformly continuous function

Question

I'm reading a proof for the claim $g(x)$ is uniformly continuous.

It comes down to: $\forall x,y>B:\left|g(x)-g(y)\right|\le \left|x-y\right| + \frac{\varepsilon}{2}$

The auther claims $\delta = \min\{\delta_1,\varepsilon/2,1\}$

• $\delta_1$ is for the closed interval $[0,B]$ (By Cantor's theorem)
• $\frac{\varepsilon}{2}$ is clear from the inequality
• What the $1$ is for?

Full question and proof:
Let $f(x)$ such that $\lim\limits_{x\to \infty} f(x)=1$ and $g(x) = f(x)\sin x$. Show that $g(x)$ is uniformly continuous.

Proof:
$g(x) = f(x)\sin x = \sin x + (f(x) -1)\sin x$.

There is $x_0$ such that for all $x>x_0$: $\left|f(x)-1\right| \le \varepsilon / 4\Rightarrow \left|(f(x)-1)\sin x \right| \le \varepsilon / 4$

$g(x)$ continuous so by Cantor's theorem it's uniformly continuous on $[0,x_0 +1]$.
Meaning, for all $\varepsilon > 0$ there is $\delta_1$ such that for all $x,y\in [0,x_0 +1]: \left| x-y \right|< \delta_1 \Rightarrow \left| f(x)-f(y) \right| < \varepsilon$

If $x,y> x_0$ then,
$\left|g(x)-g(y)\right| \le \left|\sin x - \sin y\right| + \varepsilon/4 + \varepsilon/4 \le \left|x-y\right| + \varepsilon/2$

And the rest is clear..

Answer 1

The answer of Darksonn is already pointing in the right direction.

Notice that there are two different proofs depending on whether you're working on the interval $[0,x_0+1]$ or $[x_0,\infty)$. However, for the argument of either of these to work, both $x$ and $y$ have to lie within the same interval. This you can accomplish by ensuring that $| x - y | < 1$, and hence the additional requirement that $\delta \le 1$.

Answer 2

A $1$ in a $\delta$-$\epsilon$ proof is almost always from somewhere in the proof where the author said

I would like this value (ex. $|x+1|$) to never be larger than $1$, so I can prove this much easier, and then you can assume it's not for the rest of the proof. Then at the end you have to deal with the case of it being larger than $1$ and you simply bound the $\delta$ value by $1$ to make sure that when this $|x+1|$ value is larger than $1$, the value of $\delta$ still works.