MATH 1004 Lecture 19: W-2

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14 Jan 2016

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F (x) = for x 0 for x < 0. Examples: find, if lim x 2+ x + 2 (cid:17) (cid:16)x 2. 1 x2 x g(x) = (x|x|) lim x 0+ x < 0. It is easier to remember f is continuous at x = a if limx a f (x) = f (a) Example: show f is continuous at x = 1, x = 2 if. 2x x2: properties of limits of functions: f (x) = Let f, g be two given functions and limx a f (x) and limx a g(x) both exist and are. Examples: determine whether the following limits exist. (cid:40) lim x 0 f (x) = x + 2 x x 0 x > 0. |f (x)| = 0 lim x a lim x a f (x) = 0. Example: determine discontinuity points for the following functions (cid:40) g(x) = x.

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5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

sketch the graph of a function that satisfies of the giving conditions

f''(5)=0 , f'(x)<0 when x<5

f''(x)>0 when x>5 , f''(2)=0 , f'''(7)=0

f''(x)<0 when x<2 or x>7 ,

f''(x)>0 for 2<x<7 , lim _{x row to infinity} f(x)=3 , lim _{x row to minus infinity} f(x)=3

MATH 1004 Lecture 19: W-2

MATH 1004 Full Course Notes

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