Is the proof of $\lim\limits_{x\to0} \frac{\sin(x)}{x}$ using Taylor's theorem invalid?

Question

I know that proving the $\lim\limits_{x\to0} \frac{\sin(x)}{x}$ another way is true but my question is from this proof (1) on proofwiki.

Is the method they use to proof the equation invalid? It seems to create circular logic as you can see in at the beginning of this video on youtube.

Also from the following:

Statement A: $\,\frac{d}{dx} \sin(x)\big|_{x=0} = 1$

Statement B: $\,\sin(x) = \sum_{k=0}^\infty \tfrac{(-1)^k x^{2k+1}}{(2k+1)!}$

Statement C: $\,\lim\limits_{x \to 0} \tfrac{\sin(x)}{x} = 1$

Here is the problem:

$$A\implies B\implies C\implies A$$ This is a circular.

So the method of proof (1) on the proofwiki site is a false method?

They use infinite sum which is taylor series of $\sin(x)$ which requires the derivative of $\sin(x)$ which creates circular logic?

Or it is still true provided the infinite sum of $\sin(x)$ derives from a taylor series centred at another point not $0$.

In this case can we avoid taking derivative of $\sin(x)$ at $0$ ?

So on the proof (1) they should add condition that sin(x) derive from taylor series centred not at $0$? Am I understanding this correctly?

When you take the series as a definition of $\sin$, that's taylor's theorem

"In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point."

So to avoid that should we think that statement $B$ is derived from taylor series centred not at $0$?

Please also watch the video provided in the question. It also has circular logic problem.

Please provide an easy understanding and step by step answer. I'm a newbie in calculus $2$ and have no more knowledge than calculus $2$.

Answer 1

The proof cannot be circular. One way to do all of this consistently is to define $\sin$, $\cos$ and $e^z$ by power series and then the limits, derivatives and further analytical properties follow from this definition.

However, you may ask then

How does this definition relate to the geometric meaning of $\sin$, $\cos$?

I have been wondering about this as a student and it's a pity that most books avoid this. One way to connect the geometric and algebraic definitions goes as follows:

• Using the power-series definitions, show (using elementary computations) that $e^{z_1 + z_2}=e^{z_1}e^{z_2}$ and $\overline{e^z}=e^{\overline{z}}$. As a consequence, $|e^{it}|=1$ for real $t$.
• Consider the curve $\varphi(t)=e^{it}$ and show that the derivative is $\varphi'(t)=i e^{it}$. In particular, $|\varphi'(t)|=1$ and so $\varphi$ represents a movement on the circle with unit speed. There are just two ways how such constant-speed movement is possible: either you rotate counter-clockwise, or clockwise.
• Note that the movement must be "couter-clockwise" because $\varphi'(0)=+i$.
• Define $\pi$ to be the smallest positive number for which $\varphi(2\pi)=1$ and show that $\sin$, $\cos$, $e^{it}$ are $2\pi$-periodic (using uniqueness of a solution of the differential equation $\dot{x}(t)= i x(t)$ with an initial condition.)
• The fact that $\sin$, $\cos$ are projections of $\varphi$ to the real and imaginary axis follows immediately from the power series definition.

There is also a "reversed" way: $\sin$ and $\cos$ and $\pi$ can be defined so that they satisfy a simple set of axioms (all of those reflecting the geometrical intuition) and then one proves that such functions exist, are unique, have the derivatives as required, and then the power series as a result.

Answer 2

A and C are equivalent (C is the definition of A). B implies A and C but A (or C) does not imply B. It is not stated like this on wiki, either.

Answer 3

Lookie here!

It provides geometric reasoning for why $\frac d{dx}\sin(x)=\cos(x)$. From this, you should easily see that when $x\pm h$ is in the first quadrant, then:

$$\frac{\sin(x-h)-\sin(x)}{-h}<\cos(x)<\frac{\sin(x+h)-\sin(x)}{h}$$

So the derivative follows quite clearly. One can similarly do this for $\cos$ and quite clearly see that

$$\left.\frac d{dx}\sin(x)\right|_{x=0}=\lim_{h\to0}\frac{\sin (h)}h=1$$

And of course, Taylor's theorem follows from all of this.