We have,
$a_n = a _1 \cdot a_2 \cdot a_3 \ldots a_{n-1}+4, a_1=5$
$a_2 =a_1+4=5+4=9$
$a_3 =a_1 a_2+4=5 \times 9+4=49$
$a_4 =a_1 a_2 a_3=5 \times 9 \times 49+4=2209$
$a_4=\left(a_3-2\right)^2=(49-2)^2=47^2=2209$
$a_3 =\left(a_2-2\right)^2=(9-2)^2=49$
$a_n =\left(a_{n-1}-2\right)^2$
$\therefore \quad \lim _{n \rightarrow \infty} \frac{\sqrt{ a _n}}{a_{n-1}} =\lim _{n \rightarrow \infty} \frac{\sqrt{\left(a_{n-1}-2\right)^2}}{a_{n-1}}$
$=\lim _{n \rightarrow \infty}\left(\frac{ a _{n-1}-2}{a_{n-1}}\right)=1$
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