Question

Question #b5a4d

Answer 1

`lim_(n to infty)|a_(n+1)/a_n|=9|x-2|`

Explanation:

You are on the right track! You can simplify a little further.

`|a_(n+1)/a_n|=|-3^(2(n+1)-2n)(x-2)|cdot n/(n+1)`

`=3^(cancel(2n)+2-cancel(2n))|x-2|cdot n/(n+1)`

`=9|x-2|cdot n/(n+1)`

By taking the limit,

`lim_(n to infty)|a_(n+1)/a_n|=9|x-2|lim_(n to infty)n/(n+1)`

By dividing the numerator and the denominator by `n`,

`=9|x-2|lim_(n to infty)1/(1+1/n)=9|x-2|1/(1+0)=9|x-2|`

Can you go from here?