## Calculus 10th Edition

$\lim\limits_{x \to -\infty} \frac{4x^2+5}{x^2+3} = 4$
We only look at the terms in the numerator and denominator with highest degree, as they are the only terms with significance as x approaches infinity. $\lim\limits_{x \to -\infty}\frac{4x^2+5}{x^2+3} = \lim\limits_{x \to -\infty}\frac{4x^2}{x^2} = \lim\limits_{x \to -\infty}4 = 4$