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If I let r be a positive number, I have to show that there exists a positive number, u, such that |x| < u implies that |(sin(x))/x - 1| < r. Now I've gone about this two different ways, but hit dead ends both times.

My first approach was to use the triangle inequality and the fact that -1 <= sin(x) <= 1. |(sin(x))/x - 1| <= |(sin(x))/x| + 1 = |sin(x)|/|x| + 1 <= 1/|x| + 1. I knew that if I could show that 1/|x| + 1 < r, that would mean |(sin(x))/x - 1| < r. However, the issue is that if 1/|x| + 1 < r, than 1 < r. But I need prove it with r being any positive number.

My second approach was to split it into left-hand and right-hand limits and use the squeeze theorem. I knew that if I show that each limit was 1, then the entire limit was 1. I decided to start with the left-hand limit. For x<0, 1/x <= sin(x)/x <= -1/x. However, since the limit as x approaches 0 from the left of 1/x = -oo and the limit as x approaches 0 from the left of -1/x is oo, the squeeze theorem really can't be applied.

What is it that I'm doing wrong?

Logic
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  • I don’t believe that I see any use of the fact that $x$ is the radian measure of the angle when you’re talking about $\sin x$. – Lubin Jan 18 '18 at 21:22
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    How do you define $\sin x?$ That greatly affects the proof. – Thomas Andrews Jan 18 '18 at 21:26
  • @EthanBolker, I do not accept the proof offered in the duplicate MSE entry. – Lubin Jan 18 '18 at 21:28
  • But one way is to prove, for $0<x<\pi/2$, that $$0<\sin x < x<\tan x=\frac{\sin x}{\cos x}.$$ From this, we conclude that $\cos x <\frac{\sin x}{x}<1$ and then use the squeeze, or more directly, use that $1-\cos x = \frac{\sin^2 x}{1+\cos x}<\sin^2 x<x^2.$ – Thomas Andrews Jan 18 '18 at 21:30
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    @Lubin Why do you not accept that answer? – Thomas Andrews Jan 18 '18 at 21:40
  • I don’t accept that answer, @ThomasAndrews, because it starts with an inequality, stated but not proved, that is very nearly equivalent to the desired fact. – Lubin Jan 19 '18 at 04:12

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