Evaluate limit as x approaches infinity of $\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$ I am having trouble figuring out how to answer this question by determining the degree of the numerator and/or denominator:
$$\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$$
I have tried deriving the first coefficient of the numerator and denominator, but not sure how to proceed to find the limit as $x \to \infty$.
 A: Notice, $$\lim_{x\to \infty}\frac{\sqrt{x^3+7x}}{\sqrt{4x^3+5}}$$
$$=\lim_{x\to \infty}\sqrt{\frac{x^3+7x}{4x^3+5}}$$
$$=\lim_{x\to \infty}\sqrt{\frac{x^3\left(1+\frac{7}{x^2}\right)}{x^3\left(4+\frac{5}{x^3}\right)}}$$
$$=\lim_{x\to \infty}\sqrt{\frac{1+\frac{7}{x^2}}{4+\frac{5}{x^3}}}$$
$$=\sqrt{\frac{1+0}{4+0}}=\color{red}{\frac{1}{2}}$$
A: Hint. You may write
$$
\begin{align}
\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}&=\frac{\sqrt{1 +7x^{-2}}}{\sqrt{4+5x^{-3}}}
\end{align}
$$ then it becomes easier to obtain your limit as $x \to +\infty$.
A: You factor out the dominant term from the numerator such that it cancels with the dominant term in the denominator. In this case we take out a factor of $x^3$ since $x^3$ is the highest degree or dominant term in both the numerator and denominator. 
Hence, showing all intermediate steps:
$\color{blue}{\lim_{x\rightarrow \infty}\cfrac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}=\lim_{x\rightarrow \infty}\cfrac{x^\frac{3}{2}\sqrt{1 +7x^{-2}}}{x^\frac{3}{2}\sqrt{4+5x^{-3}}}}=\lim_{x\rightarrow \infty}\color{red}{\cfrac{\sqrt{1 +7x^{-2}}}{\sqrt{4+5x^{-3}}}}=\cfrac{\sqrt{1 +7(0)}}{\sqrt{4+5(0)}}=\cfrac{1}{2}$
$\color{blue}{\mathrm{This}}$ step is easier to see if you write $$\cfrac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}= \cfrac{(x^3 +7x)^\frac{1}{2}}{({4x^3+5})^\frac{1}{2}}= \cfrac{\left(x^3(1 +\frac{7}{x^2})\right)^\frac{1}{2}}{\left({x^3(4+\frac{5}{x^3}})\right)^\frac{1}{2}}=\cfrac{x^\frac{3}{2}(1 +\frac{7}{x^2})^\frac{1}{2}}{{x^\frac{3}{2}(4+\frac{5}{x^3}})^\frac{1}{2}}=\color{red}{\cfrac{\sqrt{1 +7x^{-2}}}{\sqrt{4+5x^{-3}}}}$$
