Circular reasoning in proving $\lim_{x\to a}(\sin x) = \sin a$ I just started learning about epsilon-delta limit proofs, and I want to know how to prove using the epsilon-delta definition of a limit that $\lim_{x\to a}(\sin x) = \sin a$
I tried and failed, so I looked it up online and found the trick is to use the identity $\sin x < x$. I cannot find any proofs that do not use this identity.
I had never seen this identity before, so I searched for its proof and found this proof that uses the mean value theorem. Again, I haven't yet learnt the mean value theorem, but according to the website, it requires a continuous (and differentiable) function $f$.
But the concept of continuity is defined using the epsilon-delta limit definition! In fact, the fact that $\sin(x)$ is continuous is exactly the statement that I'm trying to prove above: $\lim_{x\to a}(\sin x) = \sin a$
This is clearly circular reasoning. My question is how does one escape it? Either there must be a way to prove $\lim_{x\to a}(\sin x) = \sin a$ without the identity $\sin x < x$, or we need to prove $\sin x < x$ without the fact that sin is continuous. Or I suppose there could be a 3rd option? I can't find any answers on how to do it, which I find most strange...
 A: A very analytic approach is to start from integrals and define $\log, \exp, \sin$ and show that these are smooth, and therefore continuous, on their domains.
First we define the natural logarithm by
$$
\ln x := \int_1^x \frac{dt}{t}
$$
It's easy to show the logarithm laws using this definition and integration rules, and that $\ln$ is differentiable.
Then we define the exponential function as its inverse,
$$
\exp := \ln^{-1}
$$
By the inverse function theorem, $\exp$ is differentiable and thus continuous.
The Maclaurin/Laurent series of $\exp$ has infinite radius of convergence so $\exp$ can be extended from $\mathbb{R}$ to a smooth function on all of $\mathbb{C}.$ We can therefore define the function $\sin$ by
$$
\sin x := \frac{\exp(ix)-\exp(-ix)}{2i}
$$
which will also be smooth and thus continuous.
A: I agree with the OP's logic that circular reasoning is involved.  In fact,
I used "Calculus 2nd Ed." by Apostol to study Calculus.  In that book, Apostol
first introduces the notion of the sine and cosine functions axiomatically,
stating that he wanted to find functions that satisfy the following 4 axioms:
(1) Sine and cosine functions are defined everywhere on the real line.
(2) Special values: $\cos 0 = \sin(\pi/2) = 1, \;cos \,\pi = -1.$
(3) $\cos(y - x) = \cos y \cos x + \sin y \sin x.$
(4) For $0 < x < \pi/2$:
$\displaystyle 0 < \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}.$
Apostol subsequently:
(a) 
Demonstrated geometrically that the traditional notion of sine and cosine
satisfies the above axioms as long as $\sin x, \,\cos x$ is interpreted as
$\sin x$ radians, $\cos x$ radians : where $2\pi$ radians = $360^{\circ}.$
(b) 
Demonstrated that all of the other trignometric identities, including those
involving continuity, differentiation, integration, and taylor series of sine and cosine
flow from these axioms.
....................
If I understand the OP's query correctly, he wants to establish that the sine
function is continuous everywhere, without the benefit of axiom (4), above.
It is very difficult for me to determine whether this is possible; it is very
difficult to precisely determine what subsequent results ulimately require
axiom (4), above.
Perhaps the real question is - what is the intended solution?  I will take a stab
at it.  In my attempt (below), I am assuming that the sine function
is continuous at $x = 0.$  It could easily be argued, in light of what the
OP is being asked to prove, that this assumption is unwarranted.
To the best of my knowledge, one of the consequences of axioms (1) through (3) above is [i.e. axiom (4) not used here]:
(5) $\displaystyle \sin x - \sin a = 2 \sin\left(\frac{x - a}{2}\right) \times 
\cos\left(\frac{x + a}{2}\right).$

Since the cosine function is a bounded function
[i.e. for all $\theta, |\cos \theta| \leq 1|$], 
(5) implies that
$|\sin x - \sin a| \leq 2 \left|\sin\left(\frac{x - a}{2}\right)\right|.$
It seems to me that the assignment is to show that for all $\epsilon > 0,$
there exists a $\delta > 0$ such that 
$0 < |x - a| < \delta \Rightarrow |(\sin x) - (\sin a)| < \epsilon.$
Choose $\delta > 0$ so that $\sin (\delta/2) < \epsilon/2.$ 
Since the the sine function is assumed continuous at $x = 0,$ this means that 
(6) if $|(x - a)| < \delta,$ then $|\sin\left(\frac{x - a}{2}\right)| < \sin (\delta/2) < \epsilon/2$ 
as required.
Addendum
In reviewing my attempt above, I noticed that I forgot to add that (to the best of my knowledge) one of the other consequences of axioms (1) through (3) [i.e. axiom (4) again unused] is that 
$\sin^2(\theta) + \cos^2(\theta) = 1.$
This consequence supports the assumption that the cosine function is bounded by $\pm 1.$
Addendum -2
Although the reasoning around (6) is doable, my presentation here was also careless.
Since the sine function is assumed continuous at $\theta = 0,$ there exists a neighborhood 
around $(\theta = 0)$ such that $\alpha$ inside of this neighborhood implies that 
$|\sin(\alpha/2)| < \sin(\delta/2).$
Therefore, $(x-a)$ must be constrained to be in this neighborhood.
A: We can do this with trigonometric identities. Based on the angle addition identity $$\sin(x + \delta) = \sin(x) \cos(\delta) + \cos(x) \sin(\delta),$$ it suffices to prove continuity of $\sin(x)$ and $\cos(x)$ as $x \to 0$. Not only that, but since $\sin(-x) = -\sin(x)$ and $\cos(-x) = \cos(x)$, it suffices to prove right-continuity as $x \to 0^+$.
We are also going to use the property that $\sin(x)$ is increasing on $[0, \pi/2)$ and $\cos(x)$ is decreasing on the same interval, that $\sin(0) = 0$, $\cos(0) = 1$, and that $\sin(\pi/4) = \cos(\pi/4) = \sqrt{2}/2.$
So let's say you give me a value of $\epsilon > 0$, and you want me to find a value of $\delta > 0$ so that $\sin([0, \delta)) \subseteq [0, \epsilon)$. Because $\sin(x)$ is increasing, it suffices to find $\delta$ so that $\sin(\delta) < \epsilon$. Starting from $x = \pi/4$, I can repeatedly use the half-angle identity $$\sin \left( \frac{x}{2} \right) = \sqrt{\frac{1 - \cos(x)}{2}} = \sqrt{1 - \cos^2(x)} \sqrt{\frac{1}{2(1 + \cos(x))}} = \sin(x) \sqrt{\frac{1}{2(1 + \cos(x))}}.$$
On $[0, \pi/4]$, $\sqrt{2}/2 \leq \cos(x) \leq 1$. Since $\sqrt{2}/2 > 7/18$, we have $2(1 + \cos(x)) \geq 2(1 + 7/18) = 25/9$ in this range, which implies
\begin{align*}
\sin \left( \frac{x}{2} \right) &\leq \frac{3}{5} \sin(x), \\
\sin \left( \frac{x}{2^n} \right) &\leq \left( \frac{3}{5} \right)^n \sin(x), \
\end{align*}
for any $x \in [0, \pi/4]$. Therefore, if you tell me $\epsilon$, I can always find a value of $N$ so that $\frac{\sqrt{2}}{2}\left( \frac{3}{5} \right)^N < \epsilon$, and then my value of $\delta$ is $\frac{\pi}{4*2^N} = \frac{\pi}{2^{N+2}}$. This proves that $\lim_{x \to 0^+} \sin(x) = 0$.
Because $\cos(x) = \sqrt{1 - \sin^2(x)}$, it also follows that $$\lim_{x \to 0^+} \cos(x) = \lim_{x \to 0^+} \sqrt{1 - \sin^2(x)} = \sqrt{1 - 0^2} = 1,$$ and we have the needed right-continuity of $\sin(x)$ and $\cos(x)$ at $x = 0$. QED.
A: It is really simple to justify that $sin$ is continuous at zero with the definition after taking $\delta = \varepsilon$. Once you did that, Formula (5) in @user2661923 (above) post will take you to a nice formal justification of continuity of $sin$ (using that $cos$ is bounded).
