Problem in proof that $ \lim_{x\to0}\frac{\sin(x)}{x}$=$1$ I'm having some confusion on understanding the proof that
$$\lim_{x\to 0}\frac{\sin x}{x}=1.$$
In the proof on khanacademy and on other various sources we always end up with the inequality
$$\cos(x)\le\frac{\sin(x)}{x}\le1\
$$
And then we apply the squeeze theorem.
However in the proof we always say that it is valid for any non zero x however if this is true how can we ever say that the inequality is less than or equal to each other my question is why isn't the inequality this instead
$$\cos(x)<\frac{\sin(x)}{x}<1
$$
No one seems to explain how they make the less than or equal to statement true so which one is correct?
 A: The answers thus far are missing an important point. It's not merely allowed to switch to the weaker, $\leq$ inequality; it is necessary to do so.
Suppose $f(x) < g(x)$ for all $x$, and that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist.
In this case it is true that $\lim_{x\to a}f(x) \leq \lim_{x\to a}g(x)$.
However, It is not necessarily true that $\lim_{x\to a}f(x) < \lim_{x\to a}g(x)$.
As an example, consider
$$f(x) = 0$$
$$g(x) = |x| \text{ (if $x\neq 0$)}$$
$$g(x) = 1 \text{ (if $x = 0$)}$$
Here, $f(x) < g(x)$ for all $x$ but $\lim_{x\to 0}f(x) = 0 = \lim_{x\to 0}g(x)$
Frequently, when taking limits on both sides of a strict inequality, we're forced to weaken the inequality from $<$ to $\leq$.
In the case of $\cos(x)<\frac{\sin(x)}{x}<1$, if we wrongly insist that strict inequality holds for the limits of the functions, we end up with $1 < \lim_{x\to 0}\frac{\sin(x)}{x} < 1$, which is clearly nonsense.
There's an exercise in Calculus by Spivak that explores some of this in more detail: chapter 5, exercise 12 (3rd ed.). I mention it here because I know @The homeschooler has access to Spivak :). Here are the relevant parts:

5-12.(a) Suppose that $f(x) \leq g(x)$ for all $x$. Prove that $\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)$, provided that these limits exist.
(c) If $f(x) < g(x)$ for all $x$, does it necessarily follow that $\lim_{x \to a} f(x) < \lim_{x \to a} g(x)$?

For part (a), you can begin by assuming $\lim_{x\to a} f(x) > \lim_{x\to a} g(x)$ and then showing how this leads to a contradiction.
Part (c) can be dealt with by giving any counter-example (the $f$ and $g$ mentioned above will work, and are from the Answer Book). Part (c) emphasizes that $\leq$ is the best we can do; we can show that $\lim_{x\to a} f(x) \not> \lim_{x\to a} g(x)$, and that is all. We cannot show strict inequality holds for the limits, even though it holds for the functions for all $x$.
Though this exercise is slightly different than the case homeschooler is asking about, the arguments are largely applicable.
Edit: I've neglected an important point myself! There is a moment when we do actively choose to weaken the inequality.
Given $\cos(x)<\frac{\sin(x)}{x}<1$ in some neighborhood around $0$, the above arguments can get us to $1\leq \lim_{x \to 0}\frac{\sin(x)}{x} \leq 1$, but only if we assume that this middle limit exists, which of course we shouldn't do because that's what we're trying to prove!
Generally speaking, if $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x)$, then $\lim_{x \to a} g(x)$ exists and is equal to the other two.
If we weaken the inequalities to read $\cos(x)\leq \frac{\sin(x)}{x} \leq 1$, we can use the above theorem to get $\lim_{x \to 0}\frac{\sin(x)}{x} = 1$, all while avoiding any assumptions about existence.
Proving this general "limits+squeeze theorem" theorem is Spivak's next exercise, 5-13.
Suppose $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x)$.
For any $\varepsilon > 0$ we can find a $\delta > 0$ such that for all $x$, if $0 < |x - a| < \delta$, then $L - \varepsilon < f(x) < L + \varepsilon$, and also $L - \varepsilon < h(x) < L + \varepsilon$.
Therefore we have
$$L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon$$
which shows $|g(x) - L| < \varepsilon$.
Anyway, sorry for the confusion and I'm guessing I've managed somehow to provide simultaneously too much and not enough in this answer.
Edit 2: reading it over again, I don't think we actually need to choose to weaken the inequalities for $\cos(x)<\frac{\sin(x)}{x}<1$. The above "limits+squeeze theorem" proof works even if we replace all the $\leq$'s with strict inequalities.
I'm relieved. Choosing to weaken it is too clever for me.
In summary,
(1) You can take limits of both sides of an inequality if you know the limits exist, but doing so turns strict $<$ into $\leq$. This is because...
(2) Two functions that are unequal everywhere can still have equal limits.
(3) If you have some function sandwiched between two other functions and those functions have the same limit, then so does the middle one.
(4) The requirement "...if $a \leq b$..." is certainly met by $a <b$.
A: $a<b$ is obviously not the same as $a \leq b$.  However, if we know that $a<b$, then it is of course true that $a \leq b$.
A: $$a\le b$$ means $$a<b\text{ or }a=b.$$
It suffices that one of the conditions be true for the whole expression to be true.
You seem to interpret that $a\ne b$ invalidates it, but this is wrong.
