limx→1 nxn+1−(n+1)xn+1ex−esin⁡πx, where n = 100, is equal to

I=limx→1 nxn(x−1)−xn−1ex−esin⁡πput x = 1 + h so that as x→1, h→0. Therefore,I=limh→0 h⋅n(1+h)n−(1+h)n−1eeh−1sin⁡πh=−limx→1 n⋅h1+nc1h+nc2h2+c3h3+…πeh2ch−1h−1+nc1h+nc2h2+nc3h3+…−1sin⁡πhπh=−n2−nc2πe=−2n2−n(n−1)2πe=n2+n2(πe)=−n(n+1)2(π)if n = 100, then I=−5050πe.

limx→1 nxn+1−(n+1)xn+1ex−esin⁡πx, where n = 100, is equal to

I=limx→1 nxn(x−1)−xn−1ex−esin⁡πput x = 1 + h so that as x→1, h→0. Therefore,I=limh→0 h⋅n(1+h)n−(1+h)n−1eeh−1sin⁡πh=−limx→1 n⋅h1+nc1h+nc2h2+c3h3+…πeh2ch−1h−1+nc1h+nc2h2+nc3h3+…−1sin⁡πhπh=−n2−nc2πe=−2n2−n(n−1)2πe=n2+n2(πe)=−n(n+1)2(π)if n = 100, then I=−5050πe.

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$lim_{x \to 1} \frac{{nx}^{n + 1} - (n + 1) x^{n} + 1}{(e^{x} - e) \sin πx}$ , where n = 100, is equal to

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$\frac{100}{π e}$

- $\frac{- 5050}{π e}$

$\frac{5050}{π e}$

$\frac{- 4950}{π e}$

answer is C.

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Detailed Solution

$I = lim_{x \to 1} \frac{{nx}^{n} (x - 1) - (x^{n} - 1)}{(e^{x} - e) \sin π}$
put x = 1 + h so that as x→1, h→0. Therefore,
$\begin{array}{l} I = lim_{h \to 0} \frac{h \cdot n (1 + h)^{n} - ((1 + h)^{n} - 1)}{e (e^{h} - 1) \sin πh} \\ = - lim_{x \to 1} \frac{n \cdot h (1 +^{n} c_{1} h +^{n} c_{2} h^{2} + c_{3} h^{3} + \dots)}{πe (h^{2}) (\frac{c^{h} - 1}{h})} \\ \frac{- (1 +^{n} c_{1} h +^{n} c_{2} h^{2} +^{n} c_{3} h^{3} + \dots - 1)}{(\frac{\sin πh}{πh})} \\ = - \frac{n^{2} -^{n} c_{2}}{πe} = - [\frac{2 n^{2} - n (n - 1)}{2 πe}] = \frac{n^{2} + n}{2 (πe)} = - \frac{n (n + 1)}{2 (π)} \end{array}$
if n = 100, then $I = - (\frac{5050}{πe}) .$