Question

MyNo 4 0-2 points SCace 1.7.00 A graphing calculator is recommended. For the limit llustrate the definition by finding the largest possible values of δ that correspond toe-0.2 and e-O. (Round your answers to four decimal places.)

Answer 1

Mathematica Thu May 30, 10:44 AM Untitled-2-Wolfram Mathematica 11.3 Eile Eet sert Format Cell graphics Faluatien Palettes Windam Hep Given f(x) =x3-3x + 6 We say Limitif(x), x-2]-8 if and only if given e>0 there exists a δ>0 such that If(x)-81se whenever lx-21 PART-1 Given E-0.2 Then we have to find the maximum value of δ>0 such that lx-2<δ implies |f(x)-8kE ie we have to find the maximum value of δ>0 such that 2-δ< x < 2 +δ implies 8-0.2 < f(x) < 8+0.2 which is 7.8f(x) 8.2 Lets draw the graph of function f(x), y-7.8 and y-82 Plot [ { × ^ 3-3 * × . 6 , 7 . 8 , 8 . 2) , { , e, 4) , plotLegends-"Expressions", AxesLabel . (, "y f (r) "), or info- 8.2 125%

Mathematica Thu May 30, 10:45 AM . Untitled-2-Wolfram Mathematica 11.3 Eile Eet sert Format Cell graphics Faluatien Palettes Windam Hep We can see that there are unique values of x from x=0 to x=4 such that f(x) = 7.8 and f(x) = 8.2 Lets find these values of x uto x+1.97744)) inj71). Solve [ x ^ 3-3.x+6=8.2 &&esxs4,x) upi)-(8-2.0219 } } Hence f(1.97744)-7.8 and f(2.0219) 8.2 Therefore required value of δ is min12-1.97744, 2.0219-21 So we have, δ: 0.0219 PART -2 Given ε= 0.1 Then we have to find the maximum value of δ>0 such that lx-2|<δ implies If(x)-81e ie we have to find the maximum value of δ>0 such that 2-δ< x < 2 +δ implies 8-0.1 < f(x) < 8+0.1 which is 7.9f(x) 8.1 125%

Mathematica Thu May 30, 10:45 AM . Untitled-2-Wolfram Mathematica 11.3 Eile Eet sert Format Cell graphics Faluatien Palettes Windam Hep Lets draw the graph of function f(x), y-7.9 and y-8.1 inj7rj. Plot[{x^3-3. x + 6, 7.9, 8.1), {s, e, 4} , PlotLegends → "Expressions", AxesLabel → {x, "y f (x)"}נ or 8.4 自.2 Ca-8. 7.8 We can see that there are unique values of x from x=0 to x-4 such that f(x) = 7.9 and f(x) = 8.1 Lets find these values of x. ux+1.98881]) In4- Solve[x 3-3 Cut74 x 2.01103) ) Hence f(1.98881)- 7.9 and f(2.01103) 8.1 Therefore required value of δ is min(2-1.98881, 2.01103-2) So we have, δ= 0.01103 125%