MAT 21A Lecture Notes - Lecture 3: Constant Function, Identity Function, Asymptote

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7 Oct 2017

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Suppose f is defined on an interval containing c if f(x) is as close to l as we like, whenever x is sufficiently close to c (but x =/= c) The limit x -> c f(x) = l f(h) = [(3+h)2 9] / (h) what is limit h -> 0 f(h) h =-0. 1 f(h)=5. 9 h=-0. 01 f(h)=5. 99 h=-0. 001 f(h)=5. 99 h=0. 001 f(h)=6. 001 h=0. 01 f(h)6. 01 h=0. 1 f(h)=6. 1. [(3+h)2 9] / (h) = (9 + 6h + h2 9) / (h) (6h + h2) / h = 6+h. Plugging in using calculator doesn"t always work . No, for small positive f(x) = 1, which is not close to 0. F(x) does not exist since it continues to infinity (not to scale) Jumps: if approaching c from different direction f(x) yields different values. Grows large: as x ->c , f(x) gets larger and larger close to infinity (vertical asymptote ) Oscillation: if f(x) oscillates faster and faster as x gets closer to c.

Related Questions

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.

What are the answers for all of these questions? I am confused on how to solve for these.