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

May 22, 2016

$0$

#### Explanation:

We have the limit

${\lim}_{x \rightarrow \infty} \frac{\sqrt{9 x + {x}^{2}}}{{x}^{4} + 7}$

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

$= {\lim}_{x \rightarrow \infty} \frac{\sqrt{{x}^{2} \left(\frac{9}{x} + 1\right)}}{{x}^{4} \left(1 + \frac{7}{x} ^ 4\right)}$

Note that the $\sqrt{{x}^{2}}$ can be brought from the square root as just $x$.

$= {\lim}_{x \rightarrow \infty} \frac{x \sqrt{\frac{9}{x} + 1}}{{x}^{4} \left(1 + \frac{7}{x} ^ 4\right)} = {\lim}_{x \rightarrow \infty} \frac{\sqrt{\frac{9}{x} + 1}}{{x}^{3} \left(1 + \frac{7}{x} ^ 4\right)}$

When analyzing this as it goes to infinity, we see that $\frac{9}{x}$ and $\frac{7}{x} ^ 4$ go to $0$.

$= \frac{\sqrt{0 + 1}}{\infty \left(0 + 1\right)} = \frac{1}{\infty} = 0$

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 $x \rightarrow \infty$ would have been $\infty$.