mathop {Lim}limits_{k to 0} frac{1}{k} intlim | vedclass

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(C) Let $L = \mathop {Lim}\limits_{k 	o 0} \frac{\int\limits_0^k (1 + \sin 2x)^{\frac{1}{x}} dx}{k}$.Since the limit is of the form $\frac{0}{0}$,we

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$e^2$

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(C) Let $L = \mathop {Lim}\limits_{k 	o 0} \frac{\int\limits_0^k (1 + \sin 2x)^{\frac{1}{x}} dx}{k}$.Since the limit is of the form $\frac{0}{0}$,we apply $L$'Hopital's rule by differentiating the numerator and the denominator with respect to $k$.Using the Fundamental Theorem of Calculus (Leibnitz's rule),the derivative of the numerator is $(1 + \sin 2k)^{\frac{1}{k}}$ and the derivative of the denominator is $1$.Thus,$L = \mathop {Lim}\limits_{k 	o 0} (1 + \sin 2k)^{\frac{1}{k}}$.This is of the form $1^{\infty}$,which can be evaluated as $e^{\mathop {Lim}\limits_{k 	o 0} \frac{1}{k} \ln(1 + \sin 2k)}$.Using the standard limit $\mathop {Lim}\limits_{u 	o 0} \frac{\ln(1+u)}{u} = 1$,we have $\mathop {Lim}\limits_{k 	o 0} \frac{\ln(1 + \sin 2k)}{k} = \mathop {Lim}\limits_{k 	o 0} \left( \frac{\ln(1 + \sin 2k)}{\sin 2k} \cdot \frac{\sin 2k}{2k} \cdot 2 ight) = 1 \cdot 1 \cdot 2 = 2$.Therefore,$L = e^2$.

$\mathop {Lim}\limits_{k \to 0} \frac{1}{k} \int\limits_0^k (1 + \sin 2x)^{\frac{1}{x}} dx$

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