Question

$\underset{x\to 1}{lim}\frac{n{x}^{n+1}-(n+1)x^n+1}{(e^x-e)\sin \pi x}$ where $n=100$, is equal to

• A. $\frac{5050}{\pi e}$
• B. $\frac{100}{\pi e}$
• C. $-\frac{5050}{\pi e}$
• D. $-\frac{4950}{\pi e}$

Answer 1

The correct option is C $-\frac{5050}{\pi e}$

$\underset{x\to 1}{lim}\frac{n{x}^{n+1}-(n+1)x^n+1}{(e^x-e)\sin \pi x}$ where $n=100$
$=\underset{x\to 1}{lim}\frac{100x^{101}-101x^{100}+1}{(e^x-e)\sin \pi x}$
$=\underset{x\to 1}{lim}\frac{100x^{101}-101x^{100}+1}{e(e^{x-1}-1)\sin \pi x}$
$=\underset{x\to 1}{lim}\frac{100x^{101}-101x^{100}+1}{\pi ex(x-1)\frac{(e^{x-1}-1)}{(x-1)}\frac{\sin \pi x}{\pi x}}$
$=\underset{x\to 1}{lim}\frac{100x^{101}-101x^{100}+1}{\pi ex(x-1)}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule
$=\underset{x\to 1}{lim}\frac{100\times 101x^{100}-101\times 100x^{99}}{\pi e(2x-1)}$
Again of the form $\frac{0}{0}$,

Applying L-Hospital's rule,
$=\underset{x\to 1}{lim}\frac{100\times 101\times 100x^{99}-101\times 100\times 99x^{98}}{2\pi e}$
$=-\frac{5050}{\pi e}$