MATH 140 Lecture Notes - Lecture 3: Quotient Rule, Product Rule

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harlequinclam735

9 Sep 2015

School

University of Maryland

Department

Mathematics

Course

MATH 140

Professor

Denny Gulick

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Document Summary

Math 140 lecture 3 limits (continued) To find > 0 so that if 0 < |x-3| < , then |f(x)-7| < , then |2x-4| < , then |x-3| < /2: ex 2, let g (x)={ 2x ,x <1. Show that lim x 1 g( x)=2 : we will need to look at both cases individually, case 1: let > 0. Find > 0 so that if 0 < |x-1| < and x < 1, then |2x-2| < , then. |x-1| < /2: case 2: to find 2 > 0 so that if 0 < |x-1| < 2, then |(4x-2)-x| < and x > 1, then | 4x 4| < , then |x 1| < /4: finally, let be the minimum of the two values, which is /4. =lim x a: quotient rule - lim x a lim x a, ex: (x3 3x+2) lim x 2 (4x+1) lim x 2 lim x 2 x3+lim x 2 lim.

Related Questions

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

10. Give an e-d proof that lim(-2x + 1) = -3. Solution: Pick any e > 0 and let 8=. If |x - 21 < 8, then f (x) - L = 1 - 2.0 +1-(-3) = 1 - 2.0 + 4 = -2|- 21

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.

MATH 140 Lecture Notes - Lecture 3: Quotient Rule, Product Rule

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