$\lim_{x \to 0}\frac{\sin(x)}{x}=1$ - rigorous proof I know one proof of $\lim_{x \to 0}\frac{\sin(x)}{x}=1$, the one using squeeze theorem. I have been told, however that this proof is not fully rigorous, as it uses the concept of arc length very informally. I wonder if any can provide me with a more rigorous proof of this fact. There are, of course, a several ways how to define the sine function. I don't care which one is used.
 A: Since you don't care which definition I use: let sine be the function satisfying
$$
y'' + y = 0~~~\text{with}~~~\left\{\begin{array}{l} y'(0) = 1\\y(0) = 0 \end{array}\right.
$$
Then, by L'Hoptital's Rule, we have
$$
\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{1}{1} = 1
$$
A: I wonder why the geometric proof based on arc-length (or area) is termed as non-rigorous. The concept of integration itself arose from the need to rigorously define/extend the notion of length and area to figures which were not composed of straight lines. To me the usual geometric definition of $\cos t, \sin t$ as the coordinates of a point $P$ lying on the circle $x^{2} + y^{2} = 1$ with center $O$ is the simplest one and this is the route through which most beginners learn the properties of $\sin $ and $\cos $ functions. If $A = (1, 0)$ then the number $t$ (in $\cos t, \sin t$ in above definition) represents either the arc length $AP$ or twice the area of sector $AOP$. I really doubt if there is any student whose first interaction with the symbols $\sin, \cos $ starts with $$\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots$$
It is only in later mathematical education that one encounters the definition of $\sin x$ as an infinite series or as an infinite product or as a solution to a differential equation (and perhaps many more definitions). Note that the analytical rigorous definition of trigonometric functions which is equivalent to the above geometric version is the equation $\tan^{-1}x = \int_{0}^{x}dt/(1 + t^{2})$ for all real $x$. This is similar to defining $\log x $ as $\int_{1}^{x}dt/t$
Under the geometric definition it is very easy (especially using the concept of area rather than arc-length) to show that for $x \in (0, \pi/2)$ we have $\sin x < x < \tan x$ and then we have two results coming from it (by the way, using the geometric definition the symbol $\pi$ is defined to be the area enclosed by circle $x^{2} + y^{2} = 1$):
1) $\lim_{x \to 0}\cos x = 1$
Since $\cos x$ is even function of $x$ it is sufficient to consider $x \to 0+$ and then $1 > \cos x = 1 - 2\sin^{2}(x/2) > 1 - 2(x/2)^{2} = 1 - x^{2}/2$ so that by squeeze theorem we have $\lim_{x \to 0^{+}}\cos x = 1$.
2) $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$
Clearly from $\sin x < x < \tan x$ we get $\cos x < \dfrac{\sin x}{x} < 1$ and then by squeeze theorem $\lim_{x \to 0^{+}}\dfrac{\sin x}{x} = 1$. Since $(\sin x )/x$ is even function of $x$ considering $x \to 0^{+}$ is sufficient.
A: Use L'Hopitals Rule.  The limit is of the form
$$
\lim_{x\to 0} \frac{\sin x}{x}=\frac{0}{0}.
$$
Thus we can now apply L'Hopitals rule which consists of taking the derivative of the numerator and denominator and than taking the limit, thus we obtain
$$
\lim_{x\to 0} \frac{\cos x}{1}=1.
$$
If you are looking for something more rigorous, let me know.
