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This was previously asked at Continuity of $\sin(x)/x$, however my particular question was not addressed.

Why are we not able take the Maclaurin expansion of $\sin(x)$ and simply divide by x, to yield 1 - $\frac{x^2}{3!}$ + $\frac{x^4}{5!}$ - $\frac{x^6}{7!}$... which would give us a function that is defined at x=0 to be 1 - $\frac{0}{3!}$ + $\frac{0}{5!}$ - $\frac{0}{7!}$ = 1

Are we not able to due this due to it being incorrect to algebraically manipulate infinite series, or is there some other reason?

Ariichu
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  • For pity's sake, when you write a formula, put the whole formula into MathJax, not just the parts that seem to "need" it! – David K Nov 12 '23 at 23:33
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    Welcome to Mathematics Stack Exchange. $\sin(x)/x$ is not defined at $x=0$ even if your function from the Maclaurin expansion is – J. W. Tanner Nov 12 '23 at 23:34
  • Extending the comment of @J.W.Tanner, the flaw in your analysis is that you are not allowed to divide by $~0.~$ So, the step that you take when you divide by $~x,~$ is invalid when applied to the specific value of $~x = 0.$ – user2661923 Nov 13 '23 at 00:52

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Short answer. In order to derive the Maclaurin expansion of $\sin x$ you need to know the derivative of $\sin x$ at $0$, which depends on knowing the limit at $0$ of $(\sin x)/x$ so using the expansion to find that limit is circular reasoning.

Ethan Bolker
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I would say the reason $\sin(x)/x$ is not defined at $0$ is the same as the reason that $x/x$ is not defined at $0$. It's because the function is defined by an operation that cannot be performed at $x=0$.

The Maclaurin series $1 - \dfrac{x^2}{3!} + \dfrac{x^4}{5!} - \dfrac{x^6}{7!} \pm \cdots$ works fine as the definition of a function, but in order to get it from $\sin(x)/x$, you divided the Maclaurin series for $\sin(x)$ by $x$, which is not a defined operation when $x$ is zero. Therefore the derivation of your function as a series is valid only for non-zero $x$, and defines the function only for non-zero $x$.

Of course you can easily extend the function $\sin(x)/x$ to a continuous function $f$ over all real numbers by "filling in" the function value $f(0) = 1$. That is what you do when you derive the Maclaurin series of $\sin(x)/x$ for $x \neq 0$ and then declare it to be a function defined for all real numbers. This ability to "fill in" an undefined function value was discussed at length in the answers to the linked question.

David K
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