Question

determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
lim_{x
ightarrowinfty}\frac{17x - 2x^{2}-8}{1 + 10x+21x^{2}}

Explanation:

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

When $x\to\infty$, we divide each term in the fraction $\frac{17x - 2x^{2}-8}{1 + 10x+21x^{2}}$ by $x^{2}$. So we get $\lim_{x\to\infty}\frac{\frac{17x}{x^{2}}-\frac{2x^{2}}{x^{2}}-\frac{8}{x^{2}}}{\frac{1}{x^{2}}+\frac{10x}{x^{2}}+\frac{21x^{2}}{x^{2}}}=\lim_{x\to\infty}\frac{\frac{17}{x}- 2-\frac{8}{x^{2}}}{\frac{1}{x^{2}}+\frac{10}{x}+21}$.

Step2: Evaluate the limit of each term

We know that $\lim_{x\to\infty}\frac{a}{x^{n}} = 0$ for $a$ being a constant and $n>0$. So $\lim_{x\to\infty}\frac{17}{x}=0$, $\lim_{x\to\infty}\frac{8}{x^{2}} = 0$, $\lim_{x\to\infty}\frac{1}{x^{2}}=0$ and $\lim_{x\to\infty}\frac{10}{x}=0$.
Then $\lim_{x\to\infty}\frac{\frac{17}{x}- 2-\frac{8}{x^{2}}}{\frac{1}{x^{2}}+\frac{10}{x}+21}=\frac{0 - 2-0}{0 + 0+21}$.

Answer:

$-\frac{2}{21}$