I think its he didnt eliminate the same variables
To find percent error it's the absolute value of
Actual-Experimental divided by The actual times 100
1) 6-5.43
2) 0.57/6
3) 0.095x100
4)9.5%
Answer: u r supposed to solve for substitution/add/subtract and ur answer should be (-2,9)
Step-by-step explanation:
By inspection, it's clear that the sequence must converge to
![\dfrac32](https://tex.z-dn.net/?f=%5Cdfrac32)
because
![\dfrac{3n+1}{2n-1}=\dfrac{3+\frac1n}{2-\frac1n}\approx\dfrac32](https://tex.z-dn.net/?f=%5Cdfrac%7B3n%2B1%7D%7B2n-1%7D%3D%5Cdfrac%7B3%2B%5Cfrac1n%7D%7B2-%5Cfrac1n%7D%5Capprox%5Cdfrac32)
when
![n](https://tex.z-dn.net/?f=n)
is arbitrarily large.
Now, for the limit as
![n\to\infty](https://tex.z-dn.net/?f=n%5Cto%5Cinfty)
to be equal to
![\dfrac32](https://tex.z-dn.net/?f=%5Cdfrac32)
is to say that for any
![\varepsilon>0](https://tex.z-dn.net/?f=%5Cvarepsilon%3E0)
, there exists some
![N](https://tex.z-dn.net/?f=N)
such that whenever
![n>N](https://tex.z-dn.net/?f=n%3EN)
, it follows that
![\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|](https://tex.z-dn.net/?f=%5Cleft%7C%5Cdfrac%7B3n%2B1%7D%7B2n-1%7D-%5Cdfrac32%5Cright%7C%3C%5Cvarepsilon)
From this inequality, we get
![\left|\dfrac{3n+1}{2n-1}-\dfrac32\right|=\left|\dfrac{(6n+2)-(6n-3)}{2(2n-1)}\right|=\dfrac52\dfrac1{|2n-1|}](https://tex.z-dn.net/?f=%5Cleft%7C%5Cdfrac%7B3n%2B1%7D%7B2n-1%7D-%5Cdfrac32%5Cright%7C%3D%5Cleft%7C%5Cdfrac%7B%286n%2B2%29-%286n-3%29%7D%7B2%282n-1%29%7D%5Cright%7C%3D%5Cdfrac52%5Cdfrac1%7B%7C2n-1%7C%7D%3C%5Cvarepsilon)
![\implies|2n-1|>\dfrac5{2\varepsilon}](https://tex.z-dn.net/?f=%5Cimplies%7C2n-1%7C%3E%5Cdfrac5%7B2%5Cvarepsilon%7D)
![\implies2n-1\dfrac5{2\varepsilon}](https://tex.z-dn.net/?f=%5Cimplies2n-1%3C-%5Cdfrac5%7B2%5Cvarepsilon%7D~%5Clor~2n-1%3E%5Cdfrac5%7B2%5Cvarepsilon%7D)
![\implies n\dfrac12+\dfrac5{4\varepsilon}](https://tex.z-dn.net/?f=%5Cimplies%20n%3C%5Cdfrac12-%5Cdfrac5%7B4%5Cvarepsilon%7D~%5Clor~n%3E%5Cdfrac12%2B%5Cdfrac5%7B4%5Cvarepsilon%7D)
As we're considering
![n\to\infty](https://tex.z-dn.net/?f=n%5Cto%5Cinfty)
, we can omit the first inequality.
We can then see that choosing
![N=\left\lceil\dfrac12+\dfrac5{4\varepsilon}\right\rceil](https://tex.z-dn.net/?f=N%3D%5Cleft%5Clceil%5Cdfrac12%2B%5Cdfrac5%7B4%5Cvarepsilon%7D%5Cright%5Crceil)
will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that
![N\in\mathbb N](https://tex.z-dn.net/?f=N%5Cin%5Cmathbb%20N)
.