## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}=4$$
$$A=\lim\limits_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}$$$$A=\lim\limits_{x\to\infty}\frac{4x^4+4x^2+1}{(x^2-2x+1)(x^2+x)}$$$$A=\lim\limits_{x\to\infty}\frac{4x^4+4x^2+1}{x^4-x^3-x^2+x}$$ Divide both numerator and denominator by \$x^4$$A=\lim\limits_{x\to\infty}\frac{\frac{4x^4+4x^2+1}{x^4}}{\frac{x^4-x^3-x^2+x}{x^4}}$$$$A=\lim\limits_{x\to\infty}\frac{4+\frac{4}{x^2}+\frac{1}{x^4}}{1-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}}$$$$A=\frac{\lim\limits_{x\to\infty}(4+4\times\frac{1}{x^2}+\frac{1}{x^4})}{\lim\limits_{x\to\infty}(1-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3})}$$$$A=\frac{4+4\times0+0}{1-0-0+0}$$$$A=4$$