Answer
Both techniques yield $$\lim_{x\to\infty}\frac{100x^3-3}{x^4-2}=0.$$
Work Step by Step
1) Standard method (identify the highest power of $x$ and pull it in front of the parentheses)
$$\lim_{x\to\infty}\frac{100x^3-3}{x^4-2}=\lim_{x\to\infty}\frac{x^4\left(\frac{100x^3}{x^4}-\frac{3}{x^4}\right)}{x^4\left(1-\frac{2}{x^4}\right)}=\lim_{x\to\infty}\frac{\frac{100}{x}-\frac{3}{x^4}}{1-\frac{2}{x^4}}=\left[\frac{0-0}{1-0}\right]=0.$$
2) L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to\infty}\frac{100x^3-3}{x^4-2}=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(100x^3-3)'}{(x^4-2)'}=\lim_{x\to\infty}\frac{300x^2}{4x^3}=\lim_{x\to\infty}\frac{75}{x}=\left[\frac{75}{\infty}\right]=0.$$
