mathop {Limit}limits_{x to {x_1}} ,,frac{x}{{ | vedclass

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(B) Let the given limit be $L = \mathop {Limit}\limits_{x 	o {x_1}} \frac{x \int_{x_1}^x f(t) dt}{x - x_1}$.Since the form is $\frac{0}{0}$ as $x 	o

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$x_1 f(x_1)$

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(B) Let the given limit be $L = \mathop {Limit}\limits_{x 	o {x_1}} \frac{x \int_{x_1}^x f(t) dt}{x - x_1}$.Since the form is $\frac{0}{0}$ as $x 	o x_1$,we apply $L$'$H$ôpital's rule.Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:Numerator derivative: $\frac{d}{dx} [x \int_{x_1}^x f(t) dt] = 1 \cdot \int_{x_1}^x f(t) dt + x \cdot f(x)$ (using Leibniz's rule).Denominator derivative: $\frac{d}{dx} (x - x_1) = 1$.Thus,$L = \mathop {Limit}\limits_{x 	o {x_1}} [\int_{x_1}^x f(t) dt + x f(x)]$.As $x 	o x_1$,$\int_{x_1}^{x_1} f(t) dt = 0$.Therefore,$L = 0 + x_1 f(x_1) = x_1 f(x_1)$.The correct option is $B$.

$\mathop {Limit}\limits_{x \to {x_1}} \,\,\frac{x}{{x - {x_1}}}\,\,\int\limits_{{x_1}}^x {f(t)} \, dt$ is equal to :

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