Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 4

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PROBLEM:

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


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Solution:

Divide by the highest denominator power

\begin{align*}

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

\end{align*}

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