Is $ \tan^{-1}(n x)$ (tan inverse n x)is continuous at 0 ?
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Let $$f(x)=\tan^{-1}(n x)\implies f'(x) = \frac{n}{1+n^2x^2}<n\\ \implies |f(x)|\le n|x|.$$ Or
The safty way might to use the inequality $$|\tan x| \ge |x|$$ which is true for $|x|\le \frac{\pi}{2}$ Can you take it from here?
Guy Fsone
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1If we are assuming it is differentiable at 0, we are already assuming it is continuous... – Zach Boyd Oct 06 '17 at 18:16
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You are right but still you can over com this obstacle by using this inequality $|\tan x| \ge |x|$ which is true for $|x|\le \frac{\pi}{2}$ – Guy Fsone Oct 06 '17 at 18:34
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I assume that has a geometric proof? That's probably the right way to go. – Zach Boyd Oct 06 '17 at 18:38