Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 1

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PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to \infty }\left( \frac{6x^3+4x^2+5}{8x^3+7x-3}\right)$

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Solution:

Divide by the highest denominator power.

\begin{align*} \displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\right)&=\displaystyle \lim\limits_{x\to \infty }\left(\frac{6x^3+4x^2+5}{8x^3+7x-3}\cdot \displaystyle \frac{\displaystyle \frac{1}{x^3}}{\displaystyle \frac{1}{x^3}}\right) \\ \\ & =\displaystyle \lim\limits_{x\to \infty \:}\left(\displaystyle \frac{6+\displaystyle \frac{4}{x}+\displaystyle \frac{5}{x^3}}{8+\displaystyle \frac{7}{x^2}-\frac{3}{x^3}}\right)\\ \\ & =\displaystyle \frac{\lim\limits_{x\to \infty \:}\left(6+\displaystyle \frac{4}{x}+\frac{5}{x^3}\right)}{\lim\limits_{x\to \infty \:}\left(8+\displaystyle\frac{7}{x^2}-\displaystyle \frac{3}{x^3}\right)}\\ \\ & =\displaystyle \frac{6+0+0}{8+0-0}\\ \\ & =\displaystyle \frac{6}{8}\\ \\ & =\displaystyle \frac{3}{4} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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