The equation is lim$($n$$$)(3$n$^2+4$n$+7)/(2+3$n$+4$n$^2)$ We evaluate what is given limit through evaluating: $(2+3$n$+4$n$^2)/($n$$$)(3$n$^2+4$n$+7)$ Determining the greatest power of in the numerator and denominator x$^2-3$x$+2=($x$-2)($x$-1)$ was the fundamental step. The numerator and denominator both contain the highest power of$.$ Making everything less complicated, divide each term in the denominator and numerator by n$^2$: If $(3$n$^2+4$n$+7)/(2+3$n$+4$n$^2)=((3$n$^2)/$n$^2+4$n$/$n$^2+7/$n$^2)/(2/$n$^2+3$n$/$n$^2+(4$n$^2)/$n$^2,$ Another reduced phrase for this is $(3+4/$n$+7/$n$^2)/(2/$n$^2+3/$n$+4).$ Step Two: The limit is determined as n$$$.$ The numbers $4/$n$,7/$n$^2,2/$n$^2,$ and $3/$n$^2$ approach $0$ as n$$$.$ Consequently, $(3+0+0)/(0+0+4)=3/4$ The overall reaction comes Lim$($n$$$)(3$n$^2+4$n$+7)/(2+3$n$+4$n$^2)=$