Answer
7
Work Step by Step
We have to determine $L=\lim\limits_{x \to -1} \dfrac{x^7+1}{x+1}$
Use the factorization formula:
$x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})$
for $n=7$ and $a=-1$:
$L=\lim\limits_{x \to -1} \dfrac{x^7-(-1)^7}{x+1}=\lim\limits_{x \to -1} \dfrac{(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)}{x+1}$
Simplify:
$L=\lim\limits_{x \to -1} (x^6-x^5+x^4-x^3+x^2-x+1)$
Determine the limit:
$L=(-1)^6-(-1)^5+(-1)^4-(-1)^3+(-1)^2-(-1)+1=1+1+1+1+1+1+1=7$
