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Solved in 2 steps with 2 images
- Find the slope of the tangent line to the parabola y=4x-x^2 at the point (1,3) 1) using the definition: m=lim->a f(x)-f(a)/x-a 2)using the equation: m=lim->0 f(a+h)-f(a)/hLet ƒ(x) = (x2 - 1)/( | x| - 1). Make tables of the values of ƒ at values of x that approach c = -1 from above and below. Then estimate limx→ -1 ƒ(x).Prove rigorously that lim lxl = 0.x→ 0
- Determine the value of the constant c if limx→0 4√1+cx −1 /x = 3.Let ƒ(x) = (x2 - 9)/(x + 3). Make a table of the values of ƒ at the points x = -3.1,-3.01, -3.001, and so on as far as your calculator can go.Then estimate limx→ -3 ƒ(x). What estimate do you arrive at ifyou evaluate ƒ at x = -2.9, -2.99, -2.999,c instead?(b) If we keep the first part of the hypothesis of Theorem 5.3.6(L’Hospital’s Rule) the same but we assume that lim f'(x)/g'(x) = ∞, x→a does it necessarily follow that lim f(x)/g(x)= ∞?, x→a
- Let ƒ(x) = (x2 - 9)/(x + 3). a. Make a table of the values of ƒ at the points x = -3.1, -3.01, -3.001, and so on as far as your calculator can go. Then estimate limx--> -3 ƒ(x). What estimate do you arrive at if you evaluate ƒ at x = -2.9, -2.99, -2.999,...... instead? b. Support your conclusions in part (a) by graphing ƒ near c = -3 and using Zoom and Trace to estimate y-values on the graph as x -->-3.If 1/(x^2 + 1) ≤ f(x) ≤ 1 + x^2 for all x in the interval (−1, 1), what is lim x→0 f(x)?What is the long term behavior of your hand drawn solution; that is, what is lim t-> infinity M(t) equal to?