## Calculus: Early Transcendentals (2nd Edition)

$5a^4$
We have to determine $L=\lim\limits_{x \to a} \dfrac{x^5-a^5}{x-a}$ Use the factorization formula: $x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})$ for $n=5$: $L=\lim\limits_{x \to a} \dfrac{(x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)}{x-a}$ Simplify: $L=\lim\limits_{x \to a} (x^4+x^3a+x^2a^2+xa^3+a^4)$ Determine the limit: $L=a^4+a^3(a)+a^2(a^2)+a(a^3)+a^4=a^4+a^4+a^4+a^4+a^4=5a^4$