How to solve this limit?#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))#;x>0

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Answer 1 · 2017-03-19T17:59:13

See below.

Considering #x ne 0#

If #abs x > 1#

#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))=lim_(n->oo)x^n/x^n((1/x^n+(x^2+4)))/(x(1+1/x^n)) = (x^2+4)/x#

If #abs x < 1#

#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))=1/x#

If #x = 1#

#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))=3#

If #x = -1#

#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))# does not exists

Answer 2 · 2017-03-27T18:00:31

#"The Reqd. Lim.="1/x, if 0 lt x lt 1;#
#=3, if x=1; and,#
#=x+4/x, if 1 lt x.#

#"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.!