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In Albert Fadell's book "Calculus with Analytic Geometry" an outline of the proof to show that tan x is not convergent at $x=\pi/2$ is given. The proof runs as follows, quoted verbatim: In essence the proof is that for $0 < |x - \pi/2| < \pi/6.$ We have $$|\tan x-A|=\left|\frac{\sin x}{\cos x}-A\right|\ge\left|\frac{\sin x}{\cos x}\right|-|A|>\frac{1/2}{\lvert\cos x \rvert}-|A|\ge\frac1{2|x-\pi/2|}-|A|>\epsilon$$

How does one get $\frac{1/2}{|\cos x|}$ and $\frac{1}{2|x-\pi/2|}$? Is the domain taken to be $0<\lvert x-\pi/2|<\pi/6$?

Parcly Taxel
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Raj Mohan Kovilath
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1 Answers1

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Given that $0<|x-\pi/2|<\pi/6$, $\sin x>\frac{\sqrt3}2>\frac12$ (note that $\sin(\pi/2\pm\pi/6)=\sin\frac\pi3=\frac{\sqrt3}2$). Hence $\left|\frac{\sin x}{\cos x}\right|=\frac{|\sin x|}{|\cos x|}>\frac{1/2}{|\cos x|}$.

In the same domain for $x$ we have $|\cos x|=|\sin(x-\pi/2)|\le|x-\pi/2|$ (as can be seen by plotting the graphs – $\sin x<x$ for all $x>0$ and the opposite holds for $x<0$), so $\frac{1/2}{|\cos x|}\ge\frac1{2|x-\pi/2|}$.

Parcly Taxel
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  • (Responding to your post on CURED about this question) Source is mentioned, argument is outlined in MathJax quoted verbatim, and contentious parts are highlighted with attempt(even if short) to answer it. That's good enough. I see you made an edit, but even prior to it I'd have been ok with it being answered. As it stands I'm fully good with it. Thank you for editing and clarifying the question anyway. – Sarvesh Ravichandran Iyer May 07 '21 at 17:09