Epsilon-Delta Limit Proof: Arccos(x) Inequalitiy I'm studying a Calculus proof using notes (proving that $\lim_{x \to 1} \cos(x) = \cos(1)$ from the definition of limit). 
The text says that we get from:
$\cos(1) −\epsilon < \cos(x) < \cos(1) + \epsilon$
To:
$\arccos(\cos(1) + \epsilon) < x < \arccos(\cos(1) − \epsilon).$
I don't understand why shouldn't it simply be:
$\arccos(\cos(1) - \epsilon) < x < \arccos(\cos(1) + \epsilon).$
So the question is why the terms ($\cos(1) + \epsilon$ and $\cos(1) - \epsilon$) are "swapped". I can see in other examples with the sin(x) function that this doesn't happen, when we solve by using arcsin(x).
 A: Hint. For $y \in [0,1)$, we have
$$
(\arccos(y))'=-\frac1{\sqrt{1-y^2}}<0
$$ thus the function is decreasing on this set.
A: Or using mean value theorem for $f(x) = \cos x$ on $[1,x]$  $\to |\cos x - \cos 1| = |-\sin \theta|\cdot |x-1|\leq |x-1|$. Thus for $\epsilon > 0$ given, choose $\delta = \epsilon$, then the proof follows.
A: Let $\varepsilon > 0$. Since $\cos x = \int_{t=0}^{x}-\sin t$ for all $x \neq 0$, and since 
$$
|-\int_{t=0}^{x}\sin t + \int_{t=0}^{1}\sin t| = |\int_{x}^{1}\sin t| \underset{\text{by the MVT for integrals}}{=} |\sin c||x-1| < \varepsilon
$$
if $|x-1| < \varepsilon/|\sin c|$,
we are done.
A: Without any use of calculus it is easily proved that $\arccos x$ is a decreasing function of $x$. This is simply because the inverse function $\cos x$ is also decreasing in interval $[0, \pi]$.
To see why $\cos x$ is decreasing observe that $$\cos a - \cos b = 2\sin((a + b)/2)\sin((b - a)/2)$$ and since $a, b\in [0, \pi]$ it follows that $(a + b)/2 \in [0, \pi]$ so that $\sin((a + b)/2) \geq 0$ and clearly $|(b - a)/2| \leq \pi/2$ so that $\sin ((b - a)/2)$ has the same sign as $(b - a)$. It follows that $\cos a - \cos b$ has same sign as that of $(b - a)$. Thus if $a > b$ then $\cos a < \cos b$.
We have used the elementary results from trigonometry and the fact that sign of $\sin x$ is same as that of $x$ if $|x| \leq \pi/2$.
It should now be clear that $\arccos x$ is decreasing in $[-1, 1]$ and hence if $a, b \in [-1, 1]$ with $a < b$ then $\arccos a > \arccos b$.
