$\lim_{x\to \frac{\pi}{2}} \sin(x)=1$ using the formal definition of limits

Question

$\lim_{x\to \frac{\pi}{2}} \sin(x)=1$

to problem is that:find $\theta$ such that if $| x -\frac{\pi}{2}|<\theta$,then $|\sin(x)-1|<\epsilon$

my attempt;

we know this;$|\sin(x)|\leq|x|$ for all $x$. so

$|\sin(x)-1|<\epsilon$

$\Leftrightarrow$$|\sin(x)|-1<\epsilon$ (**) (because $|a-b|\geq |a|-|b|$)

so it is enough to put $|x|-1<\epsilon$

$\Leftrightarrow$$ -\epsilon -1 < x< \epsilon +1$

$\Leftrightarrow$$ -\epsilon -1-\frac{\pi}{2} < x -\frac{\pi}{2}< \epsilon +1-\frac{\pi}{2}<\epsilon +1+\frac{\pi}{2}$

so finally $| x -\frac{\pi}{2}|<\epsilon +1+\frac{\pi}{2}$

i am not sure if my attempt is true , And I'm a little skeptical if the step (**) is correct,i mean Is there an equivalent or just implies?so this is the first question

the second question is that ;it is necessary to use the equivalent When we are looking for made an inequality that's involve $|x|$,from $|f(x)-l|<\epsilon$; i mean when we were did some algebraic manipulation on $|f(x)-l|<\epsilon$?

Answer 1

Addendum-2 added to respond to the comment question of user unknown.

---

Addendum added to respond to the comment question of user unknown.

---

$|x - \pi/2| < \delta \iff -\delta < x - \pi/2 < \delta \iff$

$\pi/2 - \delta < x < \pi/2 + \delta$.

So, either $x = \pi/2 + \alpha ~: 0 < \alpha < \delta$

or $x = \pi/2 - \alpha ~: 0 < \alpha < \delta$.

$\underline{\text{Case 1} : ~x = \pi/2 + \alpha}$
$\sin(x) = \sin(\pi/2)\cos(\alpha) + \sin(\alpha)\cos(\pi/2) = \cos(\alpha)$.

$\underline{\text{Case 2} : ~x = \pi/2 - \alpha}$
$\sin(x) = \sin(\pi/2)\cos(\alpha) - \sin(\alpha)\cos(\pi/2) = \cos(\alpha)$.

---

In either event, the expression being scrutinized is $\cos(\alpha)$, where $0 < \alpha < \delta$.

The idea is to establish a relationship between
$\delta$ and $\epsilon$
so that when $0 < |x - \pi/2| < \delta$,
you will have that $|\sin(x) - 1| < \epsilon.$

It has been established that when $|x - \pi/2| < \delta$, that there will exist an $\alpha \in (0,\delta)$ such that $\sin(x) = \cos(\alpha)$.

This implies that $|\sin(x) - 1| = 1 - \cos(\alpha)$.

So, choose $\delta = \min [ ~\text{Arccos}(1 - \epsilon), ~\pi/4 ~].$

Edit
The $\min$ function above is used to guard against some outlandish choice of $\epsilon$, such as $\epsilon = 3,$ which would render the Arccos function$(1 - \epsilon)$ undefined.

Then, $\alpha,$ being between $0$ and $\delta$ will be such that $1 > \cos(\alpha) = \sin(x) > 1 - \epsilon.$

Therefore, $\epsilon > 1 - \sin(x) = |1 - \sin(x)|,$ as required.

---

Addendum
Responding to the comment question of user unknown.

No, your work had a number of flaws, so I thought that I would simply map out the correct path for you to study. Then, you could compare my work against yours at your leisure.

Focusing on a couple of points:

we know this;$|\sin(x)|\leq|x|$ for all $x$. so

$|\sin(x)-1|<\epsilon$

$\Leftrightarrow$ $|\sin(x)|-1<\epsilon$ (**) (because $|a-b|\geq |a|-|b|$)

You have that $|\sin(x) - 1| \geq |\sin(x)| - 1.$

Therefore, if $|\sin(x) - 1| < \epsilon$
then you must have that $|\sin(x)| - 1 < \epsilon.$

However, the reverse implication does not hold.
For example, if $r \geq s$ and $t > r$, then $t > s.$
But, with $r \geq s$, if $t > s$, that does not imply that $t > r.$

---

so it is enough to put $|x|-1<\epsilon$

True, but misleading. This is a forest for the trees situation. You are considering the limit, as $x \to \pi/2$ of $\sin(x)$. While it is true that as $x \to \pi/2$, that $\sin(x)$ will approach $(1)$, $(x)$ itself will be nowhere near $(1)$, because $\pi/2$ is nowhere near $(1)$.

So, the whole idea of scrutinizing $|x - 1|$ instead of $|\sin(x) - 1|$ is not going to get anywhere.

---

so it is enough to put $|x|-1<\epsilon$

$$\Leftrightarrow -\epsilon -1 < x< \epsilon +1 \tag1 $$

$$\Leftrightarrow -\epsilon -1-\frac{\pi}{2} < x -\frac{\pi}{2}< \epsilon +1-\frac{\pi}{2}<\epsilon +1+\frac{\pi}{2} \tag2 $$

so finally $$| x -\frac{\pi}{2}|<\epsilon +1+\frac{\pi}{2} \tag3 $$

---

The problem here is that you became confused about what you are trying to prove.

You need to prove that if

$$|x - \pi/2| < \delta \tag4 $$

then

$$|\sin(x) - 1| < \epsilon. \tag5 $$

Part of the problem that you encountered is that you weren't sure how to relate $x$ to $\sin(x)$, as I did in my answer. So, you tried to deflect the issue by instead focusing on whether $|x - \pi/2| < \epsilon.$

These are the sorts of problems that are common for new students. You have to expand your intuition so that you never lose sight of the fact that the expressions in (4) and (5) are the key considerations.

So, there is no avoiding that you somehow have to relate $x$ being close to $\pi/2$ to the effect that this will have on $\sin(x)$. Here, for this specific problem, you can re-examine my answer.

However, for the deeper issue of you're being confused, you want to focus on the expressions in (4) and (5) above, as the key considerations in problems of this type.

---

Addendum-2
Responding to the comment question of user unknown.

...does that mean my answer will be correct, because in final I find $\theta = \epsilon + 1 + \pi/2?$

No, that is not what it means. Your use of the variable $(\theta)$ in place of $(x)$ is okay, as long as it is clearly understood what the goal is, in terms of the variable $\theta$. Re-expressing (4) and (5) above, in terms of the variable $(\theta)$,

You need to prove that if

$$|\theta - \pi/2| < \delta \tag6 $$

then

$$|\sin(\theta) - 1| < \epsilon. \tag7 $$

Compare (6) and (7) above with the goal of trying to prove that

$$\theta = \epsilon + 1 + \pi/2. \tag8 $$

What (8) above expresses is a combination of (6) and (7) that does not apply to the problem, at all. The problem requires that somehow (6) above be used as a premise that leads to the conclusion in (7) above.

And the implication between (6) and (7) must somehow be triggered by a relationship that you conjure between $\delta$ and $\epsilon.$

Compare (8) and (7) above. The requirement is that $\sin(\theta)$ be shown to be close to $(1)$. The requirement is not that $\theta$ be shown to be close to $\pi/2$ or close to $(\pi/2 + 1)$. Instead, the closeness of $\theta$ to $\pi/2$ is a premise that you start with.

So, the question is begged: what underlying challenge are you faced with? It isn't explicitly taught in school, but in order to learn Math, you have to stretch your intuition. You get into trouble if you try to apply memorized rules. Instead, you need to have a deeper understanding of the pertinent definitions and theorems and worked examples.

The only way that I know of to stretch your intuition is by tackling Math problems in the specific area, so that the entire topic becomes second nature to you.

Think of learning Math like learning a foreign language, like French. In attempting to converse in French, you don't want to have to think in English and then translate your sentence word for word into French.

Then, when you hear the response in French, you would have to translate the response word for word back into English. Instead, it is far better to strive to be fluent in French, so that you think in French.

The analogy is apt here. Hopefully, by attacking enough Math problems in this area, you will get to the point that you think in Math. Then, your intuition will automatically be guided by the ideas in (4) and (5) above, or alternatively, by (6) and (7) above.