Answer:
not arithmetic
Step-by-step explanation:
An arithmetic sequence has a common difference d between consecutive terms.
Given
= 2n² - 1
Substitute in n = 1, 2, 3, 4 to generate the first 4 terms of the sequence
= 2(1)² - 1 = 2 - 1 = 1
= 2(2)² - 1 = 8 - 1 = 7
= 2(3)² - 1 = 18 - 1 = 17
= 2(4)² - 1 = 32 - 1 = 31
check the difference between consecutive terms
7 - 1 = 6
17 - 7 = 10
31 - 17 = 14
The differences are not common hence not an arithmetic sequence
All circles have the same angles, just different degrees
Since the limit becomes the undetermined form
![\displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%201%7D%20%5Cdfrac%7Bx%5E3-2x%5E2%2B3x-2%7D%7B2x%5E4-3x%2B1%7D%20%5Cto%20%5Cdfrac%7B0%7D%7B0%7D%20)
it means that both polynomials have a root at
. So, we can fact both numerator and denominator:
![x^3-2x^2+3x-2 = (x-1)(x^2-x+2)](https://tex.z-dn.net/?f=%20x%5E3-2x%5E2%2B3x-2%20%3D%20%28x-1%29%28x%5E2-x%2B2%29%20)
![2x^4-3x+1 = (x-1)(2x^3+2x^2+2x-1)](https://tex.z-dn.net/?f=%202x%5E4-3x%2B1%20%3D%20%28x-1%29%282x%5E3%2B2x%5E2%2B2x-1%29%20)
So, the fraction becomes
![\dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1}](https://tex.z-dn.net/?f=%20%5Cdfrac%7B%28x-1%29%28x%5E2-x%2B2%29%7D%7B%28x-1%29%282x%5E3%2B2x%5E2%2B2x-1%29%7D%20%3D%20%5Cdfrac%7Bx%5E2-x%2B2%7D%7B2x%5E3%2B2x%5E2%2B2x-1%7D%20)
Now, as x approaches 1, you have no problems anymore:
![\displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%201%7D%20%5Cdfrac%7Bx%5E2-x%2B2%7D%7B2x%5E3%2B2x%5E2%2B2x-1%7D%20%5Cto%20%5Cdfrac%7B2%7D%7B5%7D%20)
Answer:4/3x²+8x−20
Step-by-step explanation:Combine Like Terms:
=x+
4
3
x2+7x+−20
=(
4
3
x2)+(x+7x)+(−20)
=
4
3
x2+8x+−20