Calculus Bible online by Gill G.S.

By Gill G.S.

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INTUITIVE TREATMENT AND DEFINITIONS 49 for all x such that |x − c| < 1. Secondly, for each > 0, let δ = 1. Then |f (x) − f (c)| = |k − k| = 0 < for all x such that |x − c| < 1. This completes the required proof. 11 Show that f (x) = 3x − 4 is continuous at x = 3. Let > 0 be given. Then |f (x) − f (3)| = |(3x − 4) − (5)| = |3x − 9| = 3|x − 3| < whenever |x − 3| < . 3 We define δ = . Then, it follows that 3 lim f (x) = f (3) x→3 and, hence, f is continuous at x = 3. 12 Show that f (x) = x3 is continuous at x = 2.

INTUITIVE TREATMENT AND DEFINITIONS 47 Since M/2 > 0, there exists some δ1 > 0 such that M 2 M 3M − + M < g(x) < 2 2 M 3M 0< < g(x) < 2 2 1 2 < |g(x)| M |g(x) − M | < whenever 0 < |x − c| < δ1 , whenever 0 < |x − c| < δ1 , whenever 0 < |x − c| < δ1 , whenever 0 < |x − c| < δ1 . Let > 0 be given. Let 1 = M 2 /2. Then δ > 0 such that δ < δ1 and > 0 and there exists some 1 |g(x) − M | < 1 whenever 0 < |x − c| < δ < δ1 , M − g(x) |g(x) − M | 1 1 = = − g(x) M g(x)M |g(x)|M 1 1 = · |g(x) − M | M |g(x)| 1 2 < · · 1 M M 21 = 2 M = whenever 0 < |x − c| < δ.

LIMITS AND CONTINUITY Part (ii) Since for all θ = 2nπ ± π2 , n integer, tan θ = sin θ 1 , sec θ = cos θ cos θ it follows that tan θ and sec θ are continuous functions. Part (iii) Both cot θ and csc θ are continuous as quotients of two continuous functions where the denominators are not zero for n = nπ, n integer. 1 Evaluate each of the following limits. 1. lim x→1 x2 − 1 x3 − 1 2. lim x→0 sin(2x) x 3. lim sin 5x sin 7x x→0 4. lim+ x2 1 −4 5. lim− x2 1 −4 6. lim x−2 x2 − 4 7. lim+ x−2 |x − 2| 8.

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