This is a new post on an old saw because this is one of those things where that I can see how that, all too sadly, the way in which we've structured the current maths curriculum really doesn't make it possible to do these kinds of things the justice they deserve and I think, ultimately, that is a disservice to many learners.

The truth is, this limit cannot really be given an honest proof without an honest definition of the sine function, first. And that is *not* as easy as it seems. Even if we consider the simple notion from many trigonometric treatments that the sine is equal to the "length of the opposite side of the right triangle divided by the length of its hypotenuse", that doesn't truly solve the problem because there is actually a subtle missing element and that is that sine is not a function of a "right triangle" (though you could define that if you wanted, and it'd be easy!), but of an *angle measure*. And actually parsing out what "angle measure" means, it turns out, is essentially equivalent to defining the sine function in the first place, so this approach is circular! (pun observed after writing despite not being originally intended!)

So how do we define sine, or angle measure? Unfortunately, any approach to this is such that it *must* involve calculus. This is because the angle measure we use is "smooth and steady", meaning that, basically, if we have some angle, we'd like fractionating that angle measure to fractionate the angle in the same manner as cutting up pieces of a pie: if I have an angle with the given angle measure $\theta$, then for the measure system to work I should be able to then produce an angle with measure $\frac{\theta}{n}$, should be an angle that is geometrically the $n$-section of the angle into $n$ *congruent* smaller angles that add up to the full angle.

Yet already, we can see right there that this is not trivial: consider $n = 3$. Then we have the famous "impossible" problem of "trisection of the angle" which vexed even the ancient Greeks and for which people would keep trying to pound at until Pierre Wantzel finally proved it undoable over two *thousand* years after. We are asking for a mathematical widget that can not only trisect, but 5-sect, 629-sect, etc. angles and in a *systematic* manner to boot!

Indeed, not only is the sine function *not* trivial, we could argue that even the *exponential* function is considerably easier to treat than sine, though I won't give such a treatment here.

Thus, how do we do it? Well, the key observation is that our "steady" angle measure is one which is, effectively, defined by the *arc length* of a segment of circle intercepted by the angle when drawn at the circle's center and projected outward. In particular, this should be "obvious" from the (circularly-introduced) geometric formula

$$\mbox{Length of circular arc} = r\theta$$

Since this is only a trivial multiplication, all the nontriviality must be in either defining $\theta$ in terms of geometric angles formed by lines, or in terms of defining the "length of a circular arc" and, moreover, these two problems must be equally hard. Hence, we will begin with the arc question first and one will see that this answer will end up using a fair bit of Calculus II material to answer this Calculus I-level question about a *supposedly* pre-Calculus mathematical object. Indeed, this is what the whole "radian measure" is: it's a measure of angles in terms of the arc length of the piece they cut from a unit circle (i.e. $r = 1$). "Degrees", are then just a weird multiple unit of actual length, equal to $\frac{2\pi}{360}$ (or better, $\frac{\tau}{360}$) of some other unit length.

If you use a *somewhat* more honest Trigonometry book, you will see something to the effect that sine and cosine are defined as basically being the coordinates on a unit circle when an angle measure $\theta$ has been emplaced from the $x$-axis:

$$C(\theta) := (\cos(\theta), \sin(\theta))$$

Now as said above, $\theta$ *is* arc length. Thus, what we have above is something called an *arc length parameterization* of the circle - and that tells us how we need to proceed. First, we need a separate definition of the arc length of a circle.

How do we get that? Well, we will obviously need a more elementary circle equation, first, than the one we just gave, and that means going to the simple *algebraic* definition,

$$x^2 + y^2 = 1$$

so that now we can solve using good ole' Algebra for $x$ and $y$ in at least a semicircle:

$$x(y) = \sqrt{1 - y^2}$$
$$y(x) = \sqrt{1 - x^2}$$

And now this is where we then must introduce Calculus II-level concept - namely, integration for arc length. The arc length swept between two values of the $x$-coordinate, for a curve given with $y$ as a function of $x$, is

$$\mbox{Arc Length}(x_1, x_2) := \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$

Hence for the circle, now $\frac{dy}{dx} = \frac{1}{2} (1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$, so

$$\mbox{Arc Length}(x_1, x_2) = \int_{x_1}^{x_2} \sqrt{1 + \frac{x^2}{1 - x^2}} dx$$

which simplifies to

$$\mbox{Arc Length}(x_1, x_2) = \int_{x_1}^{x_2} \frac{1}{\sqrt{1 - x^2}} dx$$

Now, we use the Fundamental Theorem of Calculus to *define* the *inverse* sine as

$$\arcsin(x) := \int_{0}^{x} \frac{1}{\sqrt{1 - \xi^2}} d\xi$$

which is the *arc length in terms of coordinate*, and now the sine is its inverse, *coordinate in terms of arc length*:

$$\sin(\theta) := \arcsin^{-1}(\theta)$$

.

Finally, at this point, with a full, airtight definition of $\sin(x)$ now in hand, we are ready to evaluate the limit:

$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$$

Since the "real", or base, function here is really the *inverse* function, i.e. $\arcsin$, we first proceed by making a change of variables: we consider instead the limit in terms of $y$ where $y(x) := \arcsin(x)$. Note that, trivially, $\arcsin(0) = 0$ from the integral definition, thus we get

$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = \lim_{y \rightarrow 0} \frac{y}{\arcsin(y)}$$

Now for the right-hand limit, we need only consider the behavior of $\arcsin(y)$ when $y$ is small. Since the integrand, $\frac{1}{\sqrt{1 - x^2}}$, is differentiable at $x = 0$, it can be approximated with its tangent line (which really, should be also how we *define* tangent lines in the first place, as a "best approximation", a notion that can be done in an *airtight*, *intuitive* fashion through the use of a "zoom-in") and so likewise, the integral over a tiny sliver by integral of that same tangent line. By the power rule and chain rule,

$$\frac{d}{dx} \frac{1}{\sqrt{1 - x^2}} = \frac{d}{dx} (1 - x^2)^{-1/2} = \left(-\frac{1}{2}\right) (1 - x^2)^{-3/2} \cdot (-2x) = x(1 - x^2)^{3/2}$$

so the derivative at $x = 0$ is *zero* and the tangent line is horizontal: since also $\frac{1}{\sqrt{1 - x^2}}$ evaluated at $x = 0$ is $1$, the tangent is

$$T(x) := 1$$

hence

$$\int_{0}^{y} \frac{1}{\sqrt{1 - \xi^2}}\ d\xi \approx \int_{0}^{y} 1\ d\xi$$

when $y \approx 0$, and then the right-hand integral is approximately $y$, hence $\arcsin(y) \approx y$ when $y \approx 0$ and

$$\lim_{y \rightarrow 0} \frac{y}{\arcsin(y)} = \lim_{y \rightarrow 0} \frac{y}{y} = \lim_{y \rightarrow 0} 1 = 1$$

hence

$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$

QED.

Nonetheless, as I mentioned before, this doesn't solve the requirements of the question which, while I'm sure its original asker has long moved on is, nonetheless, still relevant to calculus student after calculus student up to today: prove the limit using only Calculus I/pre-Calculus methods. What I am saying is that, in fact, that is not truly honestly possible and reveals a weakness of the curriculum in that it doesn't actually follow the proper logical buildup of the mathematical edifice.

What *really* should be done is to *leave trig for later*, that is, *skip* trig and go for Calculus first. When I studied maths on my own, I did just that. In fact, I'd say, as many educators have suggested, that most people don't need either, but really need more statistics instead. Then for those who *do* pursue higher maths, if we've done algebra and statistics, we already have right there a *lot* of interesting material we can build on for calculus, including the exponential function. There is no need to add trig functions to "sweeten the mix" when it's already plenty sweet with integrals of algebraic functions like $x \mapsto \sqrt{1 - x^2}$ which is a very nice example of the area-integration relation, and such can, if emphasized more heavily, potentially invite more nuanced thinking about integrals beside just "plugging and chugging integration rules". In particular, with a more limited set of functions, we can think about other ways we might approach them like just that and/or a variety of ways to interpret the integral which can only be good, I'd think, to develop more creative thinking about problems and less drilling in methods with little real understanding gained (and rote crunching integrals is even less relevant now with computer algebra software; more important is really being able to *understand* a problem and how its parts fit together and lead to a solution. That said, rote crunching is not something I suggest *banning* either but I suggest that ideas, concepts, and creativity should come first, *then* you get into those techniques because very often they are also still useful in analysis and being fluent at them can also make you able to solve problems more quickly, e.g. you don't want to be hitting up your calculator for 2+3 all the time in grade school and you don't want to be hitting up your CAS all the time to integrate $x \mapsto x^2$). On top of that, we have effectively now *two* different functions - logarithm *and* trigonometry - which we define by integrals, which means also that we can consider that we don't have to stop there, and this exposes the artificiality of the sacredness of so-called "elementary functions" and alows us to also perhaps consider a few more artifices of that sort like $\mathrm{erf}(x)$ which seems not too much more difficult but instead we simply reply that

$$\int e^{-x^2}\ dx$$

"can't be done" which, in light of having seen such things early on, feels like another sore cheat/blemish on the curriculum.

And to finish it all off - if you say calc can't be done before trig, I'd say that too bad Archimedes isn't here, as he would probably not have shared your sentiment since in fact he was one of the earliest to develop even a *partial* concept of integration and not only that but one of his applications of it was precisely the delineation of the arc length of a circle: that is why $\pi$ is called Archimedes' constant.