Answer
$$\lim_{x\to\infty}\frac{x^4+x^3}{12x^3+128}=\infty$$
Work Step by Step
*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=a\lim_{x\to\pm\infty}\frac{1}{x^n}=a\times0=0$
$$A=\lim_{x\to\infty}\frac{x^4+x^3}{12x^3+128}$$
Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^3$ in this case:
$$A=\lim_{x\to\infty}\frac{x+1}{12+\frac{128}{x^3}}$$
$$A=\frac{\lim_{x\to\infty}(x+1)}{12+0}=\frac{\lim_{x\to\infty}(x+1)}{12}$$
As $x\to\infty$, $(x+1)$ approaches $\infty$ as well. Therefore, $$A=\frac{\infty}{12}=\infty$$
