mathop {lim }limits_{x to 1} frac{{1 + log x | vedclass

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(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 1$.Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 1$.Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 1} \frac{\frac{d}{dx}(1 + \log x - x)}{\frac{d}{dx}(1 - 2x + x^2)} = \mathop {\lim }\limits_{x 	o 1} \frac{\frac{1}{x} - 1}{-2 + 2x}$$= \mathop {\lim }\limits_{x 	o 1} \frac{1 - x}{x(-2 + 2x)} = \mathop {\lim }\limits_{x 	o 1} \frac{-(x - 1)}{2x(x - 1)}$$= \mathop {\lim }\limits_{x 	o 1} \frac{-1}{2x} = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{1 + \log x - x}}{{1 - 2x + {x^2}}} = $

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