Given-f-x-nx-n-1-n-1-x-n-1-x-p-1-x-p-x-1-x-R-and-n-p-N-N-a-Calculate-lim-x-f-x-b-Show-that-lim-x-1-n-n-1-2p-

Question Number 99713 by Ar Brandon last updated on 22/Jun/20

Given f (x) = \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{x^{p + 1} - x^{p} - x + 1}, x \in R and (n, p) \in N^{*} \times N^{*}

a ∖ C alculate \underset{x \to + \infty}{\lim f} (x)

b ∖ Show that \lim_{x \to 1} = \frac{n (n + 1)}{2 p}

Answered by Ar Brandon last updated on 23/Jun/20

b ∖ \underset{x \to 1}{\lim f} (x) = \frac{n - (n + 1) + 1}{1 - 1 - 1 + 1} = \frac{0}{0} {Passons par la loi de LH}

\frac{\frac{d}{dx} {{nx}^{n + 1} - (n + 1) x^{n} + 1}}{\frac{d}{dx} {x^{p + 1} - x^{p} - x + 1}} = \frac{n (n + 1) x^{n} - n (n + 1) x^{n - 1}}{(p + 1) x^{p} - {px}^{p - 1} - 1} \Rightarrow \underset{x \to 1}{\lim f} (x) = \frac{0}{0}

\frac{\frac{d^{2}}{{dx}^{2}} {{nx}^{n + 1} - (n + 1) x^{n} + 1}}{\frac{d^{2}}{{dx}^{2}} {x^{p + 1} - x^{p} - x + 1}} = \frac{n^{2} (n + 1) x^{n - 1} - n (n + 1) (n - 1) x^{n - 2}}{p (p + 1) x^{p - 1} - p (p - 1) x^{p - 2}}

\underset{x \to 1}{\lim f} (x) = \frac{n^{2} (n + 1) - n (n^{2} - 1)}{p (p + 1) - p (p - 1)} = \frac{n^{3} + n^{2} - n^{3} + n}{p^{2} + p - p^{2} + p} = \frac{n (n + 1)}{2 p}

Answered by Ar Brandon last updated on 23/Jun/20

a ∖ \underset{x \to \infty}{\lim f} (x) = \lim_{x \to \infty} \frac{n - (n + 1) x^{- 1} + x^{- n - 1}}{x^{p - n} - x^{p - n - 1} - x^{- n} + x^{- n - 1}}

= \lim_{x \to \infty} {\frac{n}{x^{p - n} - x^{p - n - 1}}} = {\begin{cases} n si p = n \\ + \infty si p < n \\ 0 si p > n \end{cases}