Question

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

Answer 1

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

Answer 2

`"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.!