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Answer 1 · 2017-04-14T07:19:31

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

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?