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Reference: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? I think that all proofs above are beautiful but I cannot see why we have to complicate it. We know that for small values of $x$ that $\sin x$ is approximately equal to $x$. If now let $x$ in $\sin x$ go to zero it means that $\sin x$ goes to $x$. That is, our "error" or approximation goes to zero. We divide this by $x$ and we have $1$. Is not this valid?

Only thing is that this does not look like the typical proof.

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The purposes of the question you linked to is to precisely answer why "We know that for small values of x that sinx is approximately equal to x"? – user17762 Feb 24 '13 at 0:23
@AdamYac Circularity – Git Gud Feb 24 '13 at 0:23
My posted answer to the question you cite says essentially the same thing as what you've said. – Michael Hardy Feb 24 '13 at 0:25
Alright, thanks. I first learned about the approximation and after that, the well-known limit. Hence my question. – EricAm Feb 24 '13 at 0:33
up vote 4 down vote accepted

We don't know ahead of time that $\sin x \approx x$ for small values, or anything about the error. $|\sin x - x| \approx \frac{x^3}{6}$.

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Mathematically, the statement that "for small values of $x$, $\sin(x)$ is approximately equal to $x$" can be interpreted as $$ \lim_{x\to0}\frac{\sin(x)}{x}=1\tag{1} $$ So, given $(1)$, yes, the question of the limit is pretty senseless. However, starting from scratch, that is, just given the definition of $\sin(x)$ as the ratio of two sides of a triangle, how do we know that $(1)$ is true? That is the purpose of the proofs on the page you cite.

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Can be proven by writing the power series expansion of $\sin x$:

$$ \begin{array}{rcl} \lim\limits_{x \to 0} \dfrac{\sin x}{x} & = & \lim\limits_{x \to 0} \dfrac{x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dots}{x}\\ & = & \lim\limits_{x \to 0} \left( 1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \dfrac{x^6}{7!} + \dots \right)\\ & = & 1 \end{array} $$

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We might know that for small values of $x$ that $\sin x$ is approximately equal to $x$, from the unit circle definition, if we assumed that the length of a small arc and its height were approximately equal. Then, yes, we would know that $\sin x/x$ is approximately $1$. That would mean that the limit of $\sin x /x$, provided it exists, would also be approximately $1$. But how do we know the limit exists? How would do we know the limit is exactly $1$? That is the purpose of any of the proofs.

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An often used intuitive reason why the limit is 1 is that, as stated above, from the unit circle definition, the length of the arc of a small angle is about equal to the height if the triangle.

However, if you do a similar thing with a right angled isosceles triangle with short side 1 by putting smaller and smaller similar triangles on the hypotenuse, the sum of the lengths of the legs of the small triangles stays 2, no matter how small the triangles are and how close their right angle vertices approach the hypotenuse. (Wow - that is a mouthful.)

The main conclusions, imho, are that intuition can be led astray and that, to rigorously prove the limit exists and is 1 takes a more precise definition of sin.

Math often seems to work that way. A picture is drawn, or some plausible algebraic manipulation is done (like considering a power series beyond its radius of convergence) and something reasonable is deduced. However, to prove it in a satisfactorily rigorously way often requires replacing the original, intuitive definition with a more precise definition.

Anyway, that's how it looks to me.

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