The value of lim _{x rightarrow 2} frac{1}{x- | vedclass

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(B) Let $f(x) = \int_{2}^{x} 3 t^{2} dt$. The limit is of the form $\frac{0}{0}$ as $x \rightarrow 2$.Using $L$' Hospital's rule,we differentiate the

Vedclass · Accepted Answer

$12$

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(B) Let $f(x) = \int_{2}^{x} 3 t^{2} dt$. The limit is of the form $\frac{0}{0}$ as $x ightarrow 2$.Using $L$' Hospital's rule,we differentiate the numerator and denominator with respect to $x$:$\lim _{x ightarrow 2} \frac{\frac{d}{dx} \int_{2}^{x} 3 t^{2} dt}{\frac{d}{dx} (x-2)}$By the Leibniz Integral Rule,$\frac{d}{dx} \int_{2}^{x} 3 t^{2} dt = 3 x^{2}$.So,the limit becomes $\lim _{x ightarrow 2} \frac{3 x^{2}}{1}$.Substituting $x = 2$,we get $3 	imes (2)^{2} = 3 	imes 4 = 12$.

The value of $\lim _{x \rightarrow 2} \frac{1}{x-2} \int_{2}^{x} 3 t^{2} dt$ is

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