use the properties of limits to help decide whether each limit exists. if a limit exists, find its value…

Explanation:

Step1: Divide numerator and denominator by $x^2$

$\lim_{x\rightarrow\infty}\frac{5x^{2}+2x}{6x^{2}-3x + 1}=\lim_{x\rightarrow\infty}\frac{\frac{5x^{2}}{x^{2}}+\frac{2x}{x^{2}}}{\frac{6x^{2}}{x^{2}}-\frac{3x}{x^{2}}+\frac{1}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{5+\frac{2}{x}}{6-\frac{3}{x}+\frac{1}{x^{2}}}$

Step2: Use limit properties

We know that $\lim_{x\rightarrow\infty}\frac{1}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$. So, $\lim_{x\rightarrow\infty}\frac{5+\frac{2}{x}}{6-\frac{3}{x}+\frac{1}{x^{2}}}=\frac{\lim_{x\rightarrow\infty}(5)+\lim_{x\rightarrow\infty}\frac{2}{x}}{\lim_{x\rightarrow\infty}(6)-\lim_{x\rightarrow\infty}\frac{3}{x}+\lim_{x\rightarrow\infty}\frac{1}{x^{2}}}=\frac{5 + 0}{6-0 + 0}$

Step3: Simplify the result

$\frac{5+0}{6 - 0+0}=\frac{5}{6}$

Answer:

A. $\lim_{x\rightarrow\infty}\frac{5x^{2}+2x}{6x^{2}-3x + 1}=\frac{5}{6}$