A calculus problem by Dylan Scupin-Dursema

Dividing each term by $x^7$ gives $\lim_{x\rightarrow \infty} \dfrac{\frac{8x^7}{x^7} + \frac{4x}{x^7} + \frac{3}{x^7}}{\frac{4x^7}{x^7} + \frac{444x}{x^7} + \frac{21}{x^7}} = \lim_{x\rightarrow \infty} \dfrac{8 + \frac{4}{x^6} + \frac{3}{x^7}}{4 + \frac{444}{x^6} + \frac{21}{x^7}}$ So since ${\color{#20A900}\lim_{x\rightarrow \infty} \dfrac{1}{x^6}} = {\color{#3D99F6}\lim_{x\rightarrow \infty} \dfrac{1}{x^7}} = 0$ this implies $\begin{array}{rl} \lim_{x\rightarrow \infty} \dfrac{8 + \frac{4}{x^6} + \frac{3}{x^7}}{4 + \frac{444}{x^6} + \frac{21}{x^7}} &= \dfrac{\lim_{x\rightarrow \infty} 8 + \lim_{x\rightarrow \infty} \frac{4}{x^6} + \lim_{x\rightarrow \infty} \frac{3}{x^7}}{\lim_{x\rightarrow \infty} 4 + \lim_{x\rightarrow \infty} \frac{444}{x^6} + \lim_{x\rightarrow \infty} \frac{21}{x^7}}\\ &= \dfrac{8 + 4\cdot{\color{#20A900}\lim_{x\rightarrow \infty} \frac{1}{x^6}} + 3\cdot {\color{#3D99F6}\lim_{x\rightarrow \infty} \frac{1}{x^7}}}{4 + 444 \cdot {\color{#20A900}\lim_{x\rightarrow \infty} \frac{1}{x^6}} + 21\cdot {\color{#3D99F6}\lim_{x\rightarrow \infty} \frac{1}{x^7}}}\\ &= \dfrac{8}{4} = \boxed{2} \end{array}$