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- compute: Lim x2 + 12x -5 (x->1) Lim x+2/x2-4. (x->1) Lim x2-9/x2-x-6 (x->3)1. Is it possible that f(2) = 5 but limx→2f(x) = 7? If so, what kind of function can satisfy this?:/8/$$:$;$&:@:@/@/“/“ Part 1. What do you get in each of those limits when you try to evaluate them by plugging in a value for x: (options in pic) Part 2. Which of those limits does NOT yield an indeterminate form? (1),(2),(6),(3),(4),(5)?
- Prove that lim x 2 = 4. x→ 2Let 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.Prove that a.) lim 3x-1/x-1=1 as x --> 0 b.) lim sinx/x=1 as x --> 0