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

use the properties of limits to help decide whether each limit exists. if a limit exists, find its value.\n$$\\lim_{x \\to \\infty} \\frac{2x^{2}+2x}{7x^{2}-2x + 1}$$\nselect the correct choice below and, if necessary, fill in the answer box within your choice.\na. $$\\lim_{x \\to \\infty} \\frac{2x^{2}+2x}{7x^{2}-2x + 1}=$$ (simplify your answer. type an integer or a fraction.)\nb. the limit does not exist and is neither $$\\infty$$ nor $$-\\infty$$.

use the properties of limits to help decide whether each limit exists. if a limit exists, find its value.\n$$\\lim_{x \\to \\infty} \\frac{2x^{2}+2x}{7x^{2}-2x + 1}$$\nselect the correct choice below and, if necessary, fill in the answer box within your choice.\na. $$\\lim_{x \\to \\infty} \\frac{2x^{2}+2x}{7x^{2}-2x + 1}=$$ (simplify your answer. type an integer or a fraction.)\nb. the limit does not exist and is neither $$\\infty$$ nor $$-\\infty$$.

Answer

Explanation:

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

$$\lim_{x\rightarrow\infty}\frac{2x^{2}+2x}{7x^{2}-2x + 1}=\lim_{x\rightarrow\infty}\frac{\frac{2x^{2}}{x^{2}}+\frac{2x}{x^{2}}}{\frac{7x^{2}}{x^{2}}-\frac{2x}{x^{2}}+\frac{1}{x^{2}}}$$

Step2: Simplify the expression

$$=\lim_{x\rightarrow\infty}\frac{2+\frac{2}{x}}{7-\frac{2}{x}+\frac{1}{x^{2}}}$$

Step3: Use the limit property (\lim_{x\rightarrow\infty}\frac{c}{x^{n}} = 0) ((n>0,c) is a constant)

As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{2}{x}=0), (\lim_{x\rightarrow\infty}\frac{1}{x^{2}}=0) So (\lim_{x\rightarrow\infty}\frac{2+\frac{2}{x}}{7-\frac{2}{x}+\frac{1}{x^{2}}}=\frac{2 + 0}{7-0 + 0})

Answer:

A. (\lim_{x\rightarrow\infty}\frac{2x^{2}+2x}{7x^{2}-2x + 1}=\frac{2}{7})