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- Let h(x) = (x2 - 2x - 3)/(x2 - 4x + 3). a. Make a table of the values of h at x = 2.9, 2.99, 2.999, and so on. Then estimate limx-->3 h(x). What estimate do you arrive at if you evaluate h at x = 3.1, 3.01, 3.001,......instead? b. Support your conclusions in part (a) by graphing h near c = 3 and using Zoom and Trace to estimate y-values on the graph as x--> 3.Suppose g(x)<h(x)forx∈(0,1)∪(1,2) both limx→1g(x)andlimx→1h(x) exist. Must limx→1g(x)<limx→1h(x)? true or false1. Is it possible that f(2) = 5 but limx→2f(x) = 7? If so, what kind of function can satisfy this?
- 5. Consider the function f(x) = |x| + 2 at x = 0. Provide a table of inputs and outputs that demonstrate the limitdefinition of the derivative, numericallyLet 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
- Show that the function f(x,y)=8x^2 y subject to 3x−y=9 does not have an absolute minimum or maximum. (Hint: Solve the constraint for y and substitute into f.) Solve the constraint for y. y = ? Substitute into f. f(x,y)= ? Determine the behavior of f as x approaches −∞. limx→−∞f(x,y)= ? Determine the behavior of f as x approaches ∞. limx→∞f(x,y)= ? Does this show that f does not have an absolute maximum or minimum? 1. No 2. YesConsider g(x)=−7x3+7x2+9. a) limx→0−g(x)= b)limx→−1−g(x)= c)On what interval(s) is g(x) continuous?Lim as h approaches 0=[5e^x - 5e^(x+h)]/[3h]=
- 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.IN THE FOLLOWING PROBLEMS YOU WILL FIND THE LOMIT OF A FUNCTION NUMERICALLY. FIND THE VALUE OF THE LIMIT FROM THE TABLE OF FUNCTIONAL VALUES IN THE TABLE. IF THE LIMIT DOES NOT EXIST. STATE THIS. a) y=cosX-1 : X X -1 -0.5 -0.1 -0.01 0 0.01 0.1 0.5 cos X-1:X 0.04597 0.2448 0.0500 0.0040 UND -0.0050 -0.0500 0.2447 ---- lim cos x-1 : x -------- x--> 0 ---- lim cos x-1 : x x--> 0 ---- lim cos x-1 : x x--> 0In this example, we can combine the three simple results following the limit definition with the results in Theorem 1 to calculate the limits. We simply substitute the x- and y-values of the point being approached into the functional expression to find the limiting value.