details\nno additional details were added for this assignment.\nhw6 limits at infinity and asymptotes…

Explanation:

Step1: Divide numerator and denominator by x

Divide each term in the fraction $\frac{2 + 3x}{6-4x}$ by $x$. We get $\lim_{x\rightarrow\infty}\frac{\frac{2}{x}+3}{\frac{6}{x}-4}$.

Step2: Evaluate limits of individual terms

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{2}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{6}{x}=0$. So we have $\frac{0 + 3}{0-4}$.

Step3: Simplify the fraction

$\frac{3}{-4}=-\frac{3}{4}$.

Answer:

$-\frac{3}{4}$