In Exercises 3-6, lim f(x) = L and lim g(x) = M for some real numbers L and M. What, if anything, can you say about f(r) in each case? xc g(x) 3. L+0 and M +0 4. L= 0 and M÷0 5. L#0 and M = 0 6. L= 0 and M = 0
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- 1. Using tables of values, estimate lim 3x x→2 a. 3 b. 6 c. 2 d. 5 2. Using tables of values, estimate lim (x2+4x−3) x→1 a. 1 b. 3 c. 0 d. 2 3. Using tables of values, estimate lim (x3−2x2−3) and lim (x3−2x2−3) x→2- x→2+ a. -19 and -3, respectively b. -3 and -19, respectively c. both equal to -3 d. both equal to -19How do you prove c), i.e. the third bullet point, using lim(x ->a) [ f(x) - f(a) ] / [ x -a ] with the domain (-inf, 1)? Is a = 1?if f(5) = 4, f'(5) = 7, then lim of (f(x²+1) -4)/(x²-4) = ?? when x goes to 2
- show that : Lim h→0 h / √5h + 4 − 2 =4/5Show that |x2-4|<e , when 0<|x-2|<e(5+e)-1 and prove limx->2 x2=4 by using these inequalities.Suppose that functions g(t) and h(t) are defined for all values of t and g(0) = h(0) = 0. Can limt-->0 (g(t))/(h(t)) exist? If it does exist, must it equal zero? Give reasons for your answers
- How do we prove a theorem of the form ∀x(P(x)->Q(x))?Given that lim x→1 (4x − 3) = 1, illustrate Definition 2 by finding values of δ that correspond to ε = 0.1, ε = 0.05, and ε = 0.01. ε = 0.1 δ ≤ 1 ε = 0.05 δ ≤ 2 ε = 0.01 δ ≤ 3Given that lim x→4 (4x − 13) = 3, illustrate Definition 2 by finding values of delta that correspond to epsilon = 0.1, epsilon = 0.05, and epsilon = 0.01. epsilon = 0.1 delta≤ epsilon = 0.05 delta≤ epsilon = 0.01 delta≤
- Use the Squeeze Theorem to show that lim x → 0 (x2 cos(13?x)) = 0. Illustrate by graphing the functions f(x) = −x2, g(x) = x2 cos(13?x), and h(x) = x2 on the same screen. Let f(x) = −x2, g(x) = x2 cos(13?x), and h(x) = x2. Then ? ≤ cos(13?x) ≤ ? ⇒ ? ≤ x2 cos(13?x) ≤ ? . Since lim x→0 f(x) = lim x→0 h(x) = ? ,by the Squeeze Theorem we have lim x→0 g(x) = ? . I need helping finding the "?" values.The question asks to fill in the blank: For any positive integer n, lim as x goes to infinity of x^n / e^x =_____Each of Exercises 31–36 gives a function ƒ(x), a point c, and a positivenumber P. Find L = lim xSc ƒ(x). Then find a number d 7 0 suchthat for all x0 < x - c < d => ƒ(x) - L < P.