#"Reqd. Lim."=lim_(n to oo) {1+x^n(x^2+4)}/{x(x^n+1)}#
#=lim_(n to oo) {1+x^(n+2)+4x^n}/{x^(n+1)+x}......(1)#
#=lim_(n to oo) {x^n(1/x^n+x^2+4)}/{x^n(x+1/x^(n-1))#
#=lim_(n to oo) (1/x^n+x^2+4)/(x+1/x^(n-1))......(2).#
Now, we have to consider 3 Cases :-
Case 1 : #0 lt x lt 1.#
In this Case , we know that, as #n to oo, x^n to 0.#
And, #"by (1), :., Reqd. Lim.="{1+x^2(0)+4(o)}/{x(0)+x)=1/x.#
Case 2 : #x=1.#
#"The Reqd. Lim.="(1+1+4)/(1+1)=6/2=3.#
Case 3 : #x gt 1.#
In this Case, # because 0 lt 1/x lt 1, :. " as "n to oo, 1/x^n to 0.#
Hence, #"by (2), the Reqd. Lim."=(0+x^2+4)/{x+1/x(0)}=(x^2+4)/x, or, x+4/x.#
Enjoy Maths.!