Question

What is the limit of `sqrt(9x+x^2)/(x^4+7)` as x approaches infinity?

Answer 1

`0`

Explanation:

We have the limit

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

Factor out the largest-degreed terms from the numerator and denominator of the fraction.

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

Note that the `sqrt(x^2)` can be brought from the square root as just `x`.

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

When analyzing this as it goes to infinity, we see that `9/x` and `7/x^4` go to `0`.

`=sqrt(0+1)/(oo(0+1))=1/oo=0`

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There is also a more intuitive approach to limits of this type.

In the numerator, we have in a square root a polynomial of degree `2`. Since there is a square root, the "overpowering" force, that is, the term that grows fast and determines most the growth of the function, is reduced to `sqrt(x^2)=x`, or a degree of `1`.

In the denominator, the overpowering term is of degree `4`.

Since `x^4` grows much faster than just `x`, and `x^4` is in the denominator, the limit will be `0`. Had the rational fraction been inverted, the limit as `xrarroo` would have been `oo`.