Is this epsilon-delta proof that $\sin(x)$ is continuous circular?

Question

Prove that $\lim_{x\to a}\sin x=\sin a$, where $a$ is any real number.

Solution 13 here:

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preclimsoldirectory/PrecLimSol.html#SOLUTION13

claims to prove that $\sin(x)$ is continous, in other words: for every real $a$, $\lim_{x\to a}\sin(x)=\sin(a)$. The solution uses the Mean Value Theorem, which only works if the function in question is differentiable on some interval; in particular the function must be continous on some interval. So the solution is assuming that $\sin(x)$ is continous in order to prove that it is continous.

Is this solution bogus or am I missing something?

Note: I am not asking you to provide a correct proof. I already know of a correct proof (which does not use MVT).

Answer 1

You are right. That "proof" is totally wrong for the reason that you said. The same thing could be done with any function.

Answer 2

He's trying to prove $f(x)=\sin x$ is continuous, and within the proof he actually says "since $f$ is continuous" and draws conclusions from that and uses them in the proof.

That's as crystal-clear a case of circular reasoning as I can imagine.

Answer 3

The answers you got are excessively literal. They are at best technically correct on their own terms, and will gather some votes since it is fun to validate clever observation of errors in class material. But it is misleading to describe the solution as circular.

By "prove" the author of the web page means "find an explicit function $D(x)$ so that setting $\delta = D(\epsilon)$ satisfies the epsilon-based definition of continuity or limit". That is why the web page is titled SOLUTIONS TO LIMITS OF FUNCTIONS USING THE PRECISE DEFINITION OF LIMIT. The purpose of the page is to illustrate the technique of using the mean-value theorem to obtain epsilon-delta estimates. It is not circular, in applying that technique, to assume a function is continuous and differentiable, in order to prove concrete estimates demonstrating continuity at some given point.

You can of course get a joke proof by reciting the epsilon-delta definition of continuity and saying "whatever $\delta$ makes the continuity assumption true!". But that is not an estimate that can be computed on its own without the continuity assumptions. If you ask for such computability as part of what is to be proved, then there is no circularity.

To very specifically answer the criticism,

So the solution is assuming that $\sin (x)$ is continuous in order to prove that it is continuous.

No. The solution is assuming that $f(x)=\sin x$ is continuous, differentiable, and with a known formula for its derivative; in order to prove that $|f'(x)|$ is bounded (which is the key to getting a computable bound on $\delta$ as a function of $\epsilon$, which is what the page means by "using the precise definition of limit"). There are examples of differentiable functions with unbounded derivatives on a finite closed interval. The existence of the examples means that assuming continuous etc is not enough to get the intended conclusion.

At most there is a criticism that "using the precise definition of limit" is defined only from the context of a class, book, or web page. But it always implies computability (as in computability theory) and that can be taken as a very broad and sufficient definition for discussions like this one that parse things down to very fine details.