Answer
$3$
Work Step by Step
Given $$ \lim _{x \rightarrow \infty}\sqrt{ \frac{9 x^{3}+8x -4}{ 3-5x+x^3} } $$
Then
\begin{aligned} \lim _{x \rightarrow \infty}\sqrt{ \frac{9 x^{3}+8x -4}{ 3-5x+x^3} } &= \lim _{x \rightarrow \infty}\sqrt{ \frac{9 x^{3}/x^3+8x /x^3-4/x^3}{ 3/x^3-5x/x^3+x^3/x^3} } \ \ \text{ divide by } x^3\\
&=\sqrt{\frac{\lim _{x \rightarrow \infty}\left(9 +8 /x^2-4/x^3\right)}{\lim _{x \rightarrow \infty}\left(3/x^3-5 /x^2+1\right)} }\ \ \ \ \ \ \ \text{Limit Law 5 }\\
&=\sqrt{\frac{\lim _{x \rightarrow \infty}\left(9 +0-0\right)}{\lim _{x \rightarrow \infty}\left(0-0+1\right)} } \ \text{Theorem 4}\\
&=\sqrt{\frac{9}{1}} \\
&=3\end{aligned}
