Question

How do you find the limit of `sqrt(9x+x^2)/(x^4+7)` as x approaches `oo`?

Answer 1

Reqd. Lim.`=0`.

Explanation:

Reqd. Limit `=lim_(xrarroo)(sqrt(9x+x^2))/(x^4+7)`

`=lim_(xrarroo)(sqrt(x^2(9/x+1)))/(x^4(1+7/x^4)`

`=lim_(xrarroo)(xsqrt(9/x+1))/(x^4(1+7/x^4))`

`=lim_(xrarroo)(1/x^3)(sqrt(9/x+1))/(1+7/x^4)`

We need to recall, here, that,

as `xrarroo, 9/x=9*1/xrarr0, 1/x^3rarr0, and, 7/x^4rarr0`.

Hence, the reqd. lim.`=0*(sqrt(0+1)/(1+0))=0*1=0`.

In fact, `lim _(xrarroo)sqrt(9/x+1)`

`=sqrt{lim_(xrarroo)(9/x+1)`

`=sqrt(0+1)=1`.

The inter-changeability of the limit & sqrt. fun is because of the continuity of the sqrt. fun.

Answer 2

The limit exists, and it is zero.

Explanation:

Factor out the greatest power of `x` from both numerator and denominator:

`sqrt(9x+x^2)/(x^4+7) = sqrt(x^2(9/x+1))/(x^4(1+7/x^4)`

Since `sqrt(x^2)=|x|`, we can continue with

`sqrt(x^2(9/x+1))/(x^4(1+7/x^4)) = (|x|sqrt(9/x+1))/(x^4(1+7/x^4)`

Since `x` is approaching positive infinity, we have `|x|=x`. The next step is thus

`(|x|sqrt(9/x+1))/(x^4(1+7/x^4)) = sqrt(9/x+1)/(x^3(1+7/x^4))`

At this point, we're good to go: since `x` is approaching positive infinity, every quantity like `k/x^alpha` vanished, with `k` a real number and `alpha>0`.

Thus, the square root approaches one:

`sqrt(cancel(9/x)+1) \to sqrt(1)=1`

As for the parenthesis in the denominator, with similar claims we have

`(1+cancel(7/x^4))\to 1`

Thus, the global ratio behaves like

`1/(infty*1)\to 0`

as `x` approaches positive infinity.