\begin{align*}
\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)&=\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\
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& =\displaystyle \lim\limits_{x\to \infty \:}\left(\displaystyle \frac{6+\displaystyle \frac{4}{x}+\displaystyle \frac{5}{x^3}}{8+\displaystyle \frac{7}{x^2}-\frac{3}{x^3}}\right)\\
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& =\displaystyle \frac{\lim\limits_{x\to \infty \:}\left(6+\displaystyle \frac{4}{x}+\frac{5}{x^3}\right)}{\lim\limits_{x\to \infty \:}\left(8+\displaystyle\frac{7}{x^2}-\displaystyle \frac{3}{x^3}\right)}\\
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& =\displaystyle \frac{6+0+0}{8+0-0}\\
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& =\displaystyle \frac{6}{8}\\
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& =\displaystyle \frac{3}{4} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}