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- Now lets use a graphing calculator to get a graph of C=59(F32). By letting F=x and C=y, we obtain Figure 7.15. Pay special attention to the boundaries on x. These values were chosen so that the fraction (Maximumvalueofx)minus(Minimumvalueofx)95 Would be equal to 1. The viewing window of the graphing calculator used to produce Figure 7.15 is 95 pixels dots wide. Therefore, we use 95 as the denominator of the fraction. We chose the boundaries for y to make sure that the cursor would be visible on the screen when we looked for certain values. This was accomplished by setting the aforementioned fraction equal to 1 By moving the cursor to each of the F values, we can complete the table as follows. F 5 5 9 11 12 20 30 45 60 C 21 15 13 12 11 7 1 7 16 The C values are expressed to the nearest degree. Use your calculator and check the values in the table by using the equation C=59(F32).how did you get y^2-4xy to y-4x when there are three y's on top I input the numbers and got 0 which would make it where the limit does not exist.show that : Lim h→0 h / √5h + 4 − 2 =4/5
- 4. Letf(x) = (3x + 5 if x 6= 48 if x = 4.Find limx→4f(x).Evaluate the following: 1. lim √(6-y) - 2 y-> 2- y2 - 4 y + 4 2. lim |2t-1| - |2t+1| t-> 0- |t|3. lim ( v2 [[v]] - |v-4| ) v-> 2-4. lim [[x]] - |x| x-> 0+ [[x]] + 1If limx→0(2f(x)−15)=45limx→0(2f(x)−15)=45, then what must limx→0f(x)limx→0f(x) equal?
- Is there a value of a for which limx1 3x^2+ax+a+3/x^2+x-2 exists? Enter the value of a if such a exist otherwise enter 0.1. Using tables of values, estimate lim 3x x→2 a. 3 b. 6 c. 2 d. 5 2. Using tables of values, estimate lim (x2+4x−3) x→1 a. 1 b. 3 c. 0 d. 2 3. Using tables of values, estimate lim (x3−2x2−3) and lim (x3−2x2−3) x→2- x→2+ a. -19 and -3, respectively b. -3 and -19, respectively c. both equal to -3 d. both equal to -191. Is it possible that f(2) = 5 but limx→2f(x) = 7? If so, what kind of function can satisfy this?
- 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) =Show that |x2-4|<e , when 0<|x-2|<e(5+e)-1 and prove limx->2 x2=4 by using these inequalities.For all x ≥ 0, 4x − 9 ≤ f(x) ≤ x2 − 4x + 7. Find limx→4f(x). ⇒