Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to a}p(x) = p(a)$
We can express $p(x)$ as follows: $p(x) = b_n~x^n+b_{n-1}~x^{n-1}+...+b_1~x+b_0$ where each $b_i$ is a real number We can find $\lim\limits_{x \to a}p(x)$: $\lim\limits_{x \to a}p(x)$ $= \lim\limits_{x \to a}(b_n~x^n+b_{n-1}~x^{n-1}+...+b_1~x+b_0)$ $= \lim\limits_{x \to a}(b_n~x^n)+\lim\limits_{x \to a}(b_{n-1}~x^{n-1})+...+\lim\limits_{x \to a}(b_1~x)+\lim\limits_{x \to a}(b_0)$ $=b_n~a^n+b_{n-1}~a^{n-1}+...+b_1~a+b_0$ $= p(a)$ Thus, $\lim\limits_{x \to a}p(x) = p(a)$