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?
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