lim-x-1-p-x-x-2-x-3-x-p-x-1-

Question Number 132328 by liberty last updated on 13/Feb/21

\lim_{x \to 1} \frac{p - x - x^{2} - x^{3} - \dots - x^{p}}{x - 1} = ?

Answered by EDWIN88 last updated on 13/Feb/21

The limit is form \frac{0}{0}

L^{'} H \hat{o p i t a l} \Rightarrow L = \lim_{x \to 1} \frac{- 1 - 2 x - {3 x}^{2} - {4 x}^{3} - \dots - {px}^{p - 1}}{1}

L = - 1 - 2 - 3 - 4 - \dots - p

L = - \underset{AP}{\underset{⏟}{(1 + 2 + 3 + 4 + \dots + p)}} = - \frac{p (p + 1)}{2}

Answered by mathmax by abdo last updated on 13/Feb/21

f (x) = \frac{p - (x + x^{2} + \dots . + x^{p})}{x - 1} we do the cha 7 gement x - 1 = t (sot \to o)

\Rightarrow f (x) = f (1 + t) = \frac{p - (1 + t + {(1 + t)}^{2} + \dots . {(1 + t)}^{p})}{t} \Rightarrow

f (1 + t) \sim \frac{p - (1 + t + 1 + 2 t + \dots . + 1 + pt)}{t} = \frac{p - p - (1 + 2 + 3 \dots + p) t}{t}

= - \frac{p (p + 1)}{2} \Rightarrow \lim_{t \to 0} f (1 + t) = - \frac{p (p + 1)}{2} = \lim_{x \to 1} f (x)