## Calculus: Early Transcendentals (2nd Edition)

$na^{n-1}$
We have to determine $L=\lim\limits_{x \to a} \dfrac{x^n-a^n}{x-a}$, where $n$ is a positive integer. Use the factorization formula: $x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})$ $L=\lim\limits_{x \to a} \dfrac{(x-a)(x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})}{x-a}$ Simplify: $L=\lim\limits_{x \to a} (x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})$ Determine the limit: $L=a^{n-1}+a^{n-2}(a)+a^{n-2}(a^2)+...+a(a^{n-2})+a^{n-1}=a^{n-1}+a^{n-1}+a^{n-1}+...+a^{n-1}=na^{n-1}$