mathop {lim }limits_{x to {0^ + }} {left( {se | vedclass

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(B) Let $L = \mathop {\lim }\limits_{x 	o {0^ + }} {\left( {\sec \sqrt x } ight)^{\frac{{10}}{x}}}$.This is an indeterminate form of the type $1^\i

Vedclass · Accepted Answer

$e^{5}$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o {0^ + }} {\left( {\sec \sqrt x } ight)^{\frac{{10}}{x}}}$.This is an indeterminate form of the type $1^\infty$.We can write $L = e^{\mathop {\lim }\limits_{x 	o {0^ + }} \frac{{10}}{x} (\sec \sqrt x - 1)}$.Using the identity $\sec 	heta = \frac{1}{\cos 	heta}$,we have $\sec \sqrt x - 1 = \frac{1 - \cos \sqrt x}{\cos \sqrt x}$.So,$L = e^{\mathop {\lim }\limits_{x 	o {0^ + }} \frac{10}{x} \cdot \frac{1 - \cos \sqrt x}{\cos \sqrt x}}$.Using the limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,let $	heta = \sqrt x$. As $x 	o 0^+$,$	heta 	o 0^+$.Then $\mathop {\lim }\limits_{x 	o {0^ + }} \frac{1 - \cos \sqrt x}{x} = \mathop {\lim }\limits_{	heta 	o 0^+} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$.Substituting this back,$L = e^{10 \cdot \frac{1}{2} \cdot \frac{1}{\cos(0)}} = e^{5 \cdot 1} = e^5$.

$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sec \sqrt x } \right)^{\frac{{10}}{x}}}$ is equal to

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