By Gill G.S.

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1. 9 Consider f (x) = as x tends to 3. x2 − 9 In this case f is not a rational still, the problem at x = 3 is √ √ function; caused by the common factor ( x − 3). √ graph √ √ x− 3 f (x) = x2 − 9 √ √ ( x − 3) √ √ √ = √ (x + 3)( x − 3)( x + 3) 1 √ . = √ (x + 3)( x + 3) √ As x tends to 3, the reduced form of f tends to 1/(12 3). Thus, lim+ f (x) = lim− f (x) = lim f (x) = x→3 x→3 x→3 1 √ . 12 3 1 The discontinuity of f at x = 3 is removed by defining f (3) = √ . The 12 3 √ other discontinuities of f at x = −3 and x = − 3 are essential discontinuities and cannot be removed.

1. 2 If f and g are two functions that are continuous on a common domain D, then the sum, f + g, the difference, f − g and the product, f g, are continuous on D. Also, f /g is continuous at each point x in D such that g(x) = 0. Proof. If f and g are continuous at c, then f (c) and g(c) are real numbers and lim f (x) = f (c), lim g(x) = g(c). 1, we get lim(f (x) + g(x)) = lim f (x) + lim g(x) = f (c) + g(c) x→c x→c x→c lim(f (x) − g(x)) = lim f (x) − lim g(x) = f (c) − g(c) x→c x→c x→c lim(f (x)g(x)) = lim f (x) lim(g(x)) = f (c)g(c) x→c lim x→c x→c f (x) g(x) x→c limx→c f (x) f (c) = = , if g(c) = 0.

Lim+ csc x 16. lim+ cot x 17. lim− cot x 18. lim sec x π+ 19. limπ sec x sin 2x + sin 3x 20. lim x→0 x 21. lim− √ x−2 22. lim+ x→4 x−4 √ x−2 23 lim x→4 x−4 24. lim x→ 2 x→0 x→ 2 x→0 x→0 x→0 x→ 2 √ x→4 x→3 x−2 x−4 x4 − 81 x2 − 9 Sketch the graph of each of the following functions. Determine all the discontinuities of these functions and classify them as (a) removable type, (b) finite jump type, (c) essential type, (d) oscillation type, or other types. 2. LINEAR FUNCTION APPROXIMATIONS 25. f (x) = 2 27.