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Find the limit in terms of m as x approaches 2 from the right then again from the left

Oh, that is actually the simpler part.

First, the limit from the right. Simply plug it in the part that defines x>=2.

lim f(x) as x->2+ = -2(2)^2 + 2m = -8 + 2m

Next, from the left.

lim f(x) as x->2- = 3(2) + m = 6 + m

Then, to establish that the limit exists, you would just have to equate the 2 answers above. m = 14 is the right answer to that.

Hope this helps!!!!!

X 3 and x 2 are first two terms of a geometric sequence find the value of x for which the series converges

Ix+3I/Ix-2I <1

-1<(x+3)/(x-2)<1 so 1) (x+3)/(x-2)-1<0 so ((x+3-x+2))/(x-2) <0 so x<2
2) (x+3)/(x-2)+1>0 so (2x+1)/(x-2)>0 x<-1/2 and x>2 so both condition are satisfied if x<-1/2

The three terms a 1 b are in arithmetic progression The three terms 1 a b are in geometric progression

In an arithmetic sequence, there is a common and constant difference between each pair of consecutive terms. Between the first two terms, this common difference equals 1-a. Between the second and third terms, the common difference equals b-1. They must be equal to each other: 1 - a = b - 1 →b = 2-a

In a geometric sequence, there is a common ratio between each pair of consecutive terms. This ratio is a/1 and b/a for the first two pairs of consecutive terms, respectively. The ratio is equal between any pair of consecutive terms, so a/1 = b/a → a² = b.

We have a system:
b = 2-a
b = a²
By substitution, 2-a = a²
a² + a - 2 = 0
(a+2)(a-1) = 0
a = -2 or 1
If a = -2, then b = 4
if a = 1, then b = 1, but this is not a valid solution since a cannot equal b.
Therefore, a = -2, b = 4