## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 3}(5x^3-3x^2+x-6)=105$
$\lim\limits_{x \to 3}(5x^3-3x^2+x-6)$ $=\lim\limits_{x \to 3}5x^3-\lim\limits_{x \to 3}3x^2+\lim\limits_{x \to 3}x-\lim\limits_{x \to 3}6$ (sum and difference law) $=5\lim\limits_{x \to 3}x^3-3\lim\limits_{x \to 3}x^2+\lim\limits_{x \to 3}x-6$ (constant multiple law and $\lim\limits_{x \to a}c=c$) $=5\times3^3-3\times3^2+3-6$ (direct substitution property) $=105$