The value oflimx - 1xn-xn-1-xn-2- -x2-x-n-x-1

The correct option is D x→ 1limx^n+x^n 1+x^n 2+.......+x^2+x n/x 1 ( 0/0form ) =x→ 1lim{ nbsp;(x^n 1)/(x 1)+(x^n 1 1)/(x 1)+.......+(x 1)/(x 1)} Using L'H ...

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Byju's Answer

Standard XI

Mathematics

L'Hospital Rule to Remove Indeterminate Form

The value ofl...

Question

$The value of l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1}$

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Solution

The correct option is A

$l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1} (\frac{0}{0} f o r m) = l i m x \to 1 {\frac{(x^{n} - 1)}{(x - 1)} + \frac{(x^{n - 1} - 1)}{(x - 1)} + . . . . . . . + \frac{(x - 1)}{(x - 1)}}$
Using L'Hospital Rule we get -
$= n + (n - 1) + . . . . . . . + 2 + 1 = \frac{n (n + 1)}{2}$

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$The value of l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1}$

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The value oflimx→1xn+xn−1+xn−2+.......+x2+x−nx−1

The correct option is A limx→1xn+xn−1+xn−2+.......+x2+x−nx−1(00form)=limx→1{(xn−1)(x−1)+(xn−1−1)(x−1)+.......+(x−1)(x−1)} Using L'Hospital Rule we get - =n+(n−1)+.......+2+1=n(n+1)2

$The value of l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1}$