Child's Play

As the limit approaches 1, the numerator and denominator are indeterminate, with the numerator yielding zero, and the denominator yielding zero as well. The problem is a candidate for applying l'Hopital's rule.

Setting up definitions for f(x) and g(x):

$\lim _{ x\to 1 } f(x)\quad =\quad \lim _{ x\to 1 } (x+{ x }^{ 2 }+{ x }^{ 3 }+...+{ x }^{ n }+n)\quad =\quad 0\\ \\ \lim _{ x\to 1 } g(x)\quad =\quad \lim _{ x\to 1 } (x-1)\quad =\quad 0\\ \\ \lim _{ x\to 1 } \frac { f(x) }{ g(x) } \quad =\quad ?\\ \\ Applying\quad l'Hopital's\quad rule,\quad we\quad find:\\ \lim _{ x\to 1 } \frac { f'(x) }{ g'(x) } \\ \\ where\\ \\ f'(x)\quad =\quad 1+2x+3{ x }^{ 2 }+...+n{ x }^{ n-1 }+0\\ \\ g'(x)\quad =\quad 1\\ \\ The\quad limit\quad then\quad reduces\quad to\quad the\quad sum\quad of\quad integers\quad to\quad n:\\ \\ \boxed { \frac { n(n+1) }{ 2 } }$