lim_(x→1) ((nx^(n+1) −(n+1)x^n +1)/((e^x −e)sin πx)) = ?


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 34982 by rahul 19 last updated on 14/May/18

$\lim_{x \to 1} \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{(e^{x} - e) \sin π x} = ?$
Commented by math khazana by abdo last updated on 15/May/18

$f o r t h i s c a s e i t s b e t t e r t o u s e h o s p i t a l t h e o r e m$ $u (x) = {n x}^{n + 1} - (n + 1) x^{n} + 1$ $v (x) = (e^{x} - e) s i n (π x)$ $u^{^{'}} (x) = n (n + 1) x^{n} - n (n + 1) x^{n - 1}$ $u^{^{″}} (x) = n^{2} (n + 1) x^{n - 1} - n (n + 1) (n - 1) x^{n - 2}$ $u^{^{″}} (1) = n^{2} (n + 1) - n (n^{2} - 1) = n^{2} + n$ $v^{^{'}} (x) = e^{x} s i n (π x) + π (e^{x} - e) c o s (π x)$ $v^{^{″}} (x) = e^{x} s i n (π x) + π e^{x} c o s (π x) + π e^{x} c o s (π x)$ $- π^{2} (e^{x} - e) s i n (π x) \Rightarrow$ $v^{^{″}} (x) = - π e - π e = - 2 π e \Rightarrow$ ${l i m}_{x \to 1} \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{(e^{x} - e) s i n (π x)} = - \frac{n^{2} + n}{2 π e} .$
Commented by rahul 19 last updated on 16/May/18
Thank You prof. Abdo��
	Answered by ajfour last updated on 14/May/18

	$= \lim_{x \to 1} \frac{n (n + 1) x^{n - 1} (x - 1)}{e^{x} \sin π x + π (e^{x} - e) \cos π x}$ $= \lim_{x \to 1} \frac{n (n + 1) x^{n - 1} (x - 1)}{e^{x} \sin π x + π e (e^{x - 1} - 1) \cos π x}$ $= \frac{n (n + 1)}{\lim_{x \to 1} \frac{e^{x} \sin π x}{x - 1} + \lim_{x \to 1} \frac{π e \cos π x (e^{x - 1} - 1)}{x - 1}}$ $= \frac{n (n + 1)}{\lim_{x \to 1} \frac{e^{x} \sin [π + π (x - 1)]}{x - 1} - π e}$ $= \frac{n (n + 1)}{- 2 π e} = \frac{- n (n + 1)}{2 π e} .$
	Commented by rahul 19 last updated on 14/May/18

	$T h a n k y o u s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18
	$=((lim)/(x→1))((nx^(n+1) −nx^n −x^n +1)/(e(e^(x−1) −1)sinΠx)) =((lim)/(x→1)) ((nx^n (x−1−(x^n −1))/(e(e^(x−1) −1)sinΠx)) t=x−1 when x→1 t→0 =((lim)/(x→1))((nx^n (x−1)−(x−1)(x^(n−1) +x^(n−2) +....+1))/(e(e^(x−1) −1)sin(Πx))) =((lim)/(x→1 ))((nx^n −(x^(n−1) +x^(n−2) +...1))/(((e(e^(x−1) −1))/(x−1))×sinΠx)) =((lim)/(x→1))((nx^n −(x^(n−1) +x^(n−2) +...+1))/(sinΠx))× ((lim)/(x→1))(1/((e(e^(x−1) −1))/(x−1)))→its value is (1/e) =(1/e)((lim)/(x→1)) ((nx^n −(x^(n−1) +x^(n−2) +...x^(n−(n−1)) +x^(n−n) ))/(sinΠx)) N_r =(1/e)((lim)/(x→1))((n^2 x^(n−1) −{(n−1)x^(n−2) +(n−2)x^(n−3) +..)/) +∣n−(n−1)^� ∣x^(n−(n−1)−1) +0} D_r =((lim)/(x→1))ΠcosΠx so D_r =−Π using l′hospital N_r =(1/e)[n^2 −{(n−1)+(n−2)+(n−3)+...+1} 1=(n−1)+(k−1)×−1 1=n−1−k+1 k=n−1 N_r =(1/e)[n^2 −((n−1)/2)(n−1+1)} =(1/e)[((2n^2 −n^2 +n)/2)] =((n^2 +n)/(2e)) so value ofgiven limit=((n^2 +n)/(−2Πe)) pls check the answer$
	$= \frac{l i m}{x \to 1} \frac{{n x}^{n + 1} - {n x}^{n} - x^{n} + 1}{e (e^{x - 1} - 1) s i n Π x}$ $= \frac{l i m}{x \to 1} \frac{{n x}^{n} (x - 1 - (x^{n} - 1)}{e (e^{x - 1} - 1) s i n Π x}$ $t = x - 1 w h e n x \to 1 t \to 0$ $= \frac{l i m}{x \to 1} \frac{{n x}^{n} (x - 1) - (x - 1) (x^{n - 1} + x^{n - 2} + . . . . + 1)}{e (e^{x - 1} - 1) s i n (Π x)}$ $= \frac{l i m}{x \to 1} \frac{{n x}^{n} - (x^{n - 1} + x^{n - 2} + . . . 1)}{\frac{e (e^{x - 1} - 1)}{x - 1} \times s i n Π x}$ $= \frac{l i m}{x \to 1} \frac{{n x}^{n} - (x^{n - 1} + x^{n - 2} + . . . + 1)}{s i n Π x} \times$ $\frac{l i m}{x \to 1} \frac{1}{\frac{e (e^{x - 1} - 1)}{x - 1}} \to i t s v a l u e i s \frac{1}{e}$ $= \frac{1}{e} \frac{l i m}{x \to 1} \frac{{n x}^{n} - (x^{n - 1} + x^{n - 2} + . . . x^{n - (n - 1)} + x^{n - n})}{s i n Π x}$ $N_{r}$ $= \frac{1}{e} \frac{l i m}{x \to 1} \frac{n^{2} x^{n - 1} - {(n - 1) x^{n - 2} + (n - 2) x^{n - 3} + . .}{}$ $Missing \left or extra \right$ $D_{r} = \frac{l i m}{x \to 1} Π c o s Π x$ $s o D_{r} = - Π$ $u s i n g l^{'} h o s p i t a l$ $N_{r} = \frac{1}{e} [n^{2} - {(n - 1) + (n - 2) + (n - 3) + . . . + 1}$ $1 = (n - 1) + (k - 1) \times - 1$ $1 = n - 1 - k + 1$ $k = n - 1$ $N_{r} = \frac{1}{e} [n^{2} - \frac{n - 1}{2} (n - 1 + 1)}$ $= \frac{1}{e} [\frac{2 n^{2} - n^{2} + n}{2}]$ $= \frac{n^{2} + n}{2 e}$ $s o v a l u e o f g i v e n l i m i t = \frac{n^{2} + n}{- 2 Π e}$ $p l s c h e c k t h e a n s w e r$
	Commented by rahul 19 last updated on 14/May/18

	$A b s o l u t e l y c o r r e c t!$ $T h a n k y o u s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum