Let p = mathop {lim }limits_{x to 0^+} (1 + t | vedclass

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(A) Given $p = \mathop {\lim }\limits_{x 	o 0^+} (1 + 	an^2 \sqrt{x})^{\frac{1}{2x}}$.This is of the form $1^\infty$,so we use the formula $\mathop

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$\frac{1}{2}$

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(A) Given $p = \mathop {\lim }\limits_{x 	o 0^+} (1 + 	an^2 \sqrt{x})^{\frac{1}{2x}}$.This is of the form $1^\infty$,so we use the formula $\mathop {\lim }\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x 	o a} (f(x)-1)g(x)}$.$p = e^{\mathop {\lim }\limits_{x 	o 0^+} (	an^2 \sqrt{x}) \cdot \frac{1}{2x}}$.Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{	an 	heta}{	heta} = 1$,we have $\mathop {\lim }\limits_{x 	o 0^+} \frac{	an^2 \sqrt{x}}{x} = \mathop {\lim }\limits_{x 	o 0^+} \left( \frac{	an \sqrt{x}}{\sqrt{x}} ight)^2 = 1^2 = 1$.Thus,$p = e^{\frac{1}{2} \cdot 1} = e^{1/2}$.Taking the natural logarithm,$\log p = \log(e^{1/2}) = \frac{1}{2}$.

Let $p = \mathop {\lim }\limits_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$,then $\log p = $ . . .

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