The radius of a circular merry-go-round is 3 meters. which measurement is closest to the area of the merry-go-round in square me
1 answer:
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You can use the root test here. The series will converge if
![L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}](https://tex.z-dn.net/?f=L%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cfrac%7B%284-x%29%5En%7D%7B4%5En%2B9%5En%7D%7D%3C1)
You have
![L=\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{(4-x)^n}{4^n+9^n}}=|4-x|\lim_{n\to\infty}\frac1{\sqrt[n]{4^n+9^n}}](https://tex.z-dn.net/?f=L%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cfrac%7B%284-x%29%5En%7D%7B4%5En%2B9%5En%7D%7D%3D%7C4-x%7C%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac1%7B%5Csqrt%5Bn%5D%7B4%5En%2B9%5En%7D%7D)
Notice that
![\dfrac1{\sqrt[n]{4^n+9^n}}=\dfrac1{\sqrt[n]{9^n}\sqrt[n]{1+\left(\frac49\right)^n}}=\dfrac1{9\sqrt[n]{1+\left(\frac49\right)^n}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5Bn%5D%7B4%5En%2B9%5En%7D%7D%3D%5Cdfrac1%7B%5Csqrt%5Bn%5D%7B9%5En%7D%5Csqrt%5Bn%5D%7B1%2B%5Cleft%28%5Cfrac49%5Cright%29%5En%7D%7D%3D%5Cdfrac1%7B9%5Csqrt%5Bn%5D%7B1%2B%5Cleft%28%5Cfrac49%5Cright%29%5En%7D%7D)
so as
, you have 
, which means you end up with
This is the interval of convergence. The radius of convergence can be determined by finding the half-length of the interval, or by solving the inequality in terms of
so that 
is the ROC. You get
You have
Notice that
so as
This is the interval of convergence. The radius of convergence can be determined by finding the half-length of the interval, or by solving the inequality in terms of