evaluate the limit $lim_{x\rightarrowinfty}\frac{7x + 3}{6x^{2}-2x + 6}$

Explanation:

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

$\lim_{x\rightarrow\infty}\frac{7x + 3}{6x^{2}-2x + 6}=\lim_{x\rightarrow\infty}\frac{\frac{7x}{x^{2}}+\frac{3}{x^{2}}}{\frac{6x^{2}}{x^{2}}-\frac{2x}{x^{2}}+\frac{6}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{\frac{7}{x}+\frac{3}{x^{2}}}{6-\frac{2}{x}+\frac{6}{x^{2}}}$

Step2: Use limit rules

As $x\rightarrow\infty$, $\lim_{x\rightarrow\infty}\frac{1}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$. So, $\lim_{x\rightarrow\infty}\frac{\frac{7}{x}+\frac{3}{x^{2}}}{6-\frac{2}{x}+\frac{6}{x^{2}}}=\frac{0 + 0}{6-0 + 0}$

Answer:

$0$