1. ƒ(x) = {*4+ x> 2 x < 2 Evaluate lim ƒ(x) x2 а. + b. lim ƒ(x) x2 lim ƒ(x) C. ƒ(x) = {*2 x < 1 x 2 1 Evaluate а. a. lim ƒ(x) x-1+
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- 9. Given that limx→3 f(x) = 2 and limx→3 g(x) = −5 compute the following limits:(a) limx→32f(x) = (b) limx→3f(x) − 4g(x) = (c) limx→3g(x) + 3−2 + 3f(x) =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 falseIf lim f(x) =5 x-->1- and lim f(x)=5 x-->1+, what can you say about lim f(x) x-->1? What can you say about f(1)?
- Suppose that limx→2[f(x)+x2]=α. Find limx→1(x+1)f(2x) 2α−4 Not applicable 2f(2) 2(α−4)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.Determine value for c so that lim f(x) x--> 3 exists f(x)= 1/3x+c For x<3 -x+8 For x>3