Question

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

Explanation:

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

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

Step2: Simplify the expression

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

Step3: Use the limit property \(\lim_{x

ightarrow\infty}\frac{c}{x^{n}} = 0\) (\(n>0,c\) is a constant)
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{2}{x}=0\), \(\lim_{x
ightarrow\infty}\frac{1}{x^{2}}=0\)
So \(\lim_{x
ightarrow\infty}\frac{2+\frac{2}{x}}{7-\frac{2}{x}+\frac{1}{x^{2}}}=\frac{2 + 0}{7-0 + 0}\)

Answer:

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