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Was reading Thomas Calculus and came across the limit of $\frac{\sin(x)}{x}$ at $x \rightarrow 0$.

The method they used was comparing areas of sector and $2$ triangles to prove the inequality

$\sin(x) < x < \tan(x)$,

and then I remembered one of the proofs of area of circles which was done using joining $n$ triangles and so basically they used $\frac{\sin(x)}{x} \rightarrow 1$ as $x \rightarrow 0$.

Isn't it wrong to use area of circle to prove the limit?

Aris Mercier
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Macron
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    You can derive the area of a circle in other ways so the argument isn't circular. – whpowell96 Feb 02 '24 at 16:14
  • Hmmm.... wouldn't such a proof of a circle prove both $\frac {\sin x}x\to 1$ and $A = \pi r^2$ both at the same time and be a valid (for certain values of "valid") proof of both? – fleablood Feb 02 '24 at 23:45

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