Differential and Integral Calculus by Feliciano and Uy, Exercise 1.4, Problem 4 Advertisements PROBLEM: Evaluate limx→∞(x3+x+2x2−1)\displaystyle \lim\limits_{x\to \infty }\left(\frac{x^3+x+2}{x^2-1}\right)x→∞lim(x2−1x3+x+2) Advertisements Solution: Divide by the highest denominator power limx→∞(x3+x+2x2−1)=limx→∞(x3+x+2x2−1⋅1x31x3)=limx→∞(x3x3+xx3+2x3x2x3−1x3)=limx→∞(1+1x2+2x31x−1x3)=1+0+00−0=∞ (Answer)\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*}x→∞lim(x2−1x3+x+2)=x→∞lim⎝⎛x2−1x3+x+2⋅x31x31⎠⎞=x→∞lim⎝⎛x3x2−x31x3x3+x3x+x32⎠⎞=x→∞lim⎝⎛x1−x311+x21+x32⎠⎞=0−01+0+0=∞ (Answer) Advertisements Advertisements