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# solHMK8 - Section 4.1#1(a Say |x 1| < 1/6 < 1 Then 0 < x <...

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Section 4.1 #1) (a) Say | x - 1 | < 1 / 6 < 1. Then 0 < x < 2 and | x + 1 | < 3. | x 2 - 1 | = | x + 1 || x - 1 | < 3 × 1 6 = 1 2 . (b) We need | x 2 - 1 | < 1000. If | x - 1 | < 1, then 0 < x < 2 and | x + 1 | < 3. | x 2 - 1 | = | x + 1 || x - 1 | < 3 < 1000. (c) Let | x - 1 | < 1 / (3 n ) < 1. Then 0 < x < 2 and | x + 1 | < 3. | x 2 - 1 | = | x + 1 || x - 1 | < 3 × 1 3 n = 1 n . (d) Let | x - 1 | < 1 / (7 n ) < 1. Then 0 < x < 2 and | x 2 + x + 1 | ≤ | x | 2 + | x | + 1 < 7. | x 3 - 1 | = | x 2 + x + 1 || x - 1 | < 7 × 1 7 n = 1 n . #3) lim x c f ( x ) = L ⇐⇒ ( > 0)( δ > 0)0 < | x - c | < δ = ⇒ | f ( x ) - L | < ⇐⇒ ( > 0)( δ > 0)0 < | x - c | < δ = ⇒ || f ( x ) - L | - 0 | < lim x c | f ( x ) - L | = 0 #8) Let > 0. Let δ = inf { c, c } > 0. Then if | x - c | < δ , x > 0 and 1 x + sqrtc < 1 c . Therefore, | x - c | = | x - c | | x + c | < c c = . So lim x c x = c . #9) (a) Let ( x n ) be any sequence tending to 2 that is never 2 or 1. We know that lim 1 1 - x n = - 1, so it follows from the Sequential Criterion that lim x 2 1 1 - x = - 1. (b) Let ( x n ) be any sequence tending to 1 that is never 1 or - 1. We know that lim x n 1+ x n = 1 / 2, so it follows from the Sequential Criterion that lim x 1 x 1+ x = 1 / 2. (c) For x = 0, x 2 / | x | = | x | . If > 0, set δ = . Then for 0 < | x | < δ , | x | < . Hence, lim x 0 x 2 / | x | = 0.

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