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

2 Answers
Cesareo R.
Mar 19, 2017

See below.

Explanation:

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

Ratnaker Mehta
Mar 27, 2017

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

Explanation:

"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.!

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