Chapter 2 - Practice Exercises - Page 111: 42

$$\lim_{x\to-\infty}\frac{2x^2+3}{5x^2+7}=\frac{2}{5}$$

*Remember that $\lim_{x\to\pm\infty}\frac{a}{x^n}=0$ $$A=\lim_{x\to-\infty}\frac{2x^2+3}{5x^2+7}$$ Divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^2$ in this case: $$A=\lim_{x\to-\infty}\frac{2+\frac{3}{x^2}}{5+\frac{7}{x^2}}$$ $$A=\frac{2+0}{5+0}=\frac{2}{5}$$