Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 2

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PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right)$

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Solution:

Divide by the highest denominator power.

\begin{align*} \displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\right) & =\displaystyle \lim\limits_{x\to \infty }\left(\frac{3x^2+x+2}{x^3+8x+1}\cdot \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right)\\ \\ &=\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3x^2}{x^3}+\displaystyle \frac{x}{x^3}+\displaystyle \frac{2}{x^3}}{\displaystyle \frac{x^3}{x^3}+\frac{8x}{x^3}+\frac{1}{x^3}}\right)\\ \\ & =\lim\limits_{x\to \infty }\left(\displaystyle \frac{\displaystyle \frac{3}{x}+\frac{1}{x^2}+\displaystyle \frac{2}{x^3}}{1+\displaystyle \frac{8}{x^2}+\frac{1}{x^3}}\right)\\ \\ &=\frac{0+0+0}{1+0+0} \\ \\ &=0\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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