How to prove $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$ with out Squeezing?

Question

The same question have been asked here. But almost all the answers given there use the idea of squeezing one way or another, even this geometric proof uses the idea of squeezing. So, here is my question; how to prove
$$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$ without the idea of squeezing involved.

Edited: No offense, but proofs of limits using integrals or derivative(or any other concepts that are defined using limits) to me is like building a house starting from the roof. So I am not looking for those kind of proofs.

Answer 1

Use Taylor expansion of $\sin$ around zero. You have

$$\sin(x) = x+o(x)\,,$$

from which that limit follows immediately.

Answer 2

You can find the limit using geometry! It emerge when calculating the limit of the area of a regular $n$-gone inscribed on a circle of radius $r$. The limit of the area as $n\to\infty$ must be equal to the area of the circle $\pi r^2$. Imposing that and you will solve your limit. Some comments: the relevant limit that must be equal to the area of the circle is given by $$ r^2\pi\lim_{n\to\infty}\frac{\sin(2\pi/n)}{2\pi/n}. $$ Your limit is recover using the substitution $2\pi/n\mapsto x$ ($\infty\mapsto0$). The argoument of the limit follow by considering a subdivision of the $n$-gone in $n$ triangles that have the hipotenuse equal to the radius of the circle. The area of the triangle is found to be $$ A_T=r^22\frac{\cos(\pi/n)\sin(\pi/n)}{2} $$ that is equal to $$ r^2\frac{\sin(2\pi/n)}{2} $$ by the duplication formula. Clearly the area of circle $A_\mathcal{C}=\pi r^2$ is the limit of $nA_T$ when $n\to\infty$. Remember that in order to obtain the correct limit you have to multiply by $1$ written as $\pi/\pi$ and writing $n$ as $\frac{1}{1/n}$.

Answer 3

I might be wrong, but I wager that starting from geometric definition of sine, what you are asking is impossible. The reasoning is simple. Although very easy to visualize, we actually have no clue what the value of $\sin x$ is exactly, except for some special $x$'s. It is inevitable that we approximate $\sin x$ the best we can and apply squeeze theorem to that.

Note that starting with definition $\sin x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$, there is no squeezing needed, but then the problem is trivial since by definition $\lim_{x\to 0}\frac{\sin x}{x} = \sin'(0)$.