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If a(1)=1 and a(n)+1=(4+3a(n))/(3+2a(n))...

If `a_(1)=1` and `a_(n)+1=(4+3a_(n))/(3+2a_(n)),nge1"and if" lim_(ntooo) a_(n)=a,"then find the value of a."`

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We have `a_(n)+1=(4+3a_(n))/(3+2a_(n))`
`:." "underset(ntooo)lima_(n)+1=underset(ntooo)lim(4+3an)/(3+2a_(n)`
`implies" "underset(ntooo)lima_(n)+1=(4+3underset(ntooo)lima_(n))/(3+2underset(ntooo)lima_(n))`
`a=(4+3a)/(3+2a)" "(becauseunderset(ntooo)lima_(n)+1=underset(ntooo)lima_(n)=a)`
`2a^(2)=4`
`a=sqrt(2)" "(ane-sqrt(2)"because"a_(n)gt0)`
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