This is from a calculus textbook (self-study), but my problem is not with the calculus material itself. I feel I'm missing something (obvious!) from algebra. In any case: The problem statement Find an appropriate number L (a limit), and a deleted neighborhood N of a, such that a given e > 0, L - e < f(x) < L + e when x is in N. Exercise I can solve I can solve most of the exercises, but I can't figure out some of them. Here's one I can solve: 1. f(x) = 5 - 3x, a = 2 The answer in the back of the book, which is my answer, is: L = -1, 0 < | x - 2 | < e/3 My work is, L = 5 - 3*2 = 5 - 6 = -1 L - e < 5 - 3x < L + e -1 - e < 5 - 3x < -1 + e -6 - e < -3x < -6 + e (-6 - e)/-3 > x > (-6 + e)/-3 2 + e/3 > x > 2 - e/3 e/3 > x - 2 > -e/3 0 < | x - 2 | < e/3 (to delete x=2) Exercise I can't solve (or at least can't get the answer in the back of the book): 3. f(x) = 4/x, a = 2 The answer in the back of the book is: 0 < | x - 2 | < 2e/(2 + e) Here's how far I get: L = 4/2 = 2 2 - e < 4/x < 2 + e -e < 4/x - 2 < e -e/4 < x - 1/2 < e/4 Now, here is where I'm missing something. I can either delete 1/2 (instead of 2), 0 < | x - 1/2 | < e/4 or, if I subtract 3/2 instead, -e/4 - 3/2 < x - 2 < e/4 - 3/2 I can't unify the leftmost and rightmost terms because they don't have the same absolute value. Trying really hard to get the book's answer, working from one of the above equations, I can multiply by 8/(2+e): -e/4 < x - 1/2 < e/4 (8/(2+e)) * (-e/4) < (8/(2+e)) * (x - 1/2) < (8/(2+e))*(e/4) -2e/(2+e) < (8x-4)/(2+e) < 2e/(2+e) So now I'm stuck the other way because I can't reduce (8x-4)/(2+e) to (x-2). There's more problems that seem to use the same kind of manipulation that I'm failing to find. Thanks in advance.