Let A = lim_{x rightarrow 0^{+}} left(1 + tan | vedclass

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(C) Given $A = \lim_{x ightarrow 0^{+}} \left(1 + 	an^2 \sqrt{x}ight)^{\frac{1}{2x}}$.This is of the form $1^{\infty}$.Using the formula $\lim_{x

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$\frac{1}{2}$

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(C) Given $A = \lim_{x ightarrow 0^{+}} \left(1 + 	an^2 \sqrt{x}ight)^{\frac{1}{2x}}$.This is of the form $1^{\infty}$.Using the formula $\lim_{x ightarrow a} f(x)^{g(x)} = e^{\lim_{x ightarrow a} (f(x)-1)g(x)}$,we have:$\log_{e} A = \lim_{x ightarrow 0^{+}} \left(1 + 	an^2 \sqrt{x} - 1ight) \cdot \frac{1}{2x}$$= \lim_{x ightarrow 0^{+}} \frac{	an^2 \sqrt{x}}{2x}$$= \lim_{x ightarrow 0^{+}} \frac{1}{2} \left(\frac{	an \sqrt{x}}{\sqrt{x}}ight)^2$Since $\lim_{	heta ightarrow 0} \frac{	an 	heta}{	heta} = 1$,we get:$\log_{e} A = \frac{1}{2} \cdot (1)^2 = \frac{1}{2}$.

Let $A = \lim_{x \rightarrow 0^{+}} \left(1 + \tan^2 \sqrt{x}\right)^{\frac{1}{2x}}$,then $\log_{e} A = $

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