If alpha = lim_{x rightarrow 0^{+}} left( fra | vedclass

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(B) First,evaluate $\alpha$: $\alpha = \lim_{x ightarrow 0^{+}} \frac{e^{\sqrt{	an x}} - e^{\sqrt{x}}}{\sqrt{	an x} - \sqrt{x}}$. Let $u = \sqrt{\

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$6$

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(B) First,evaluate $\alpha$: $\alpha = \lim_{x ightarrow 0^{+}} \frac{e^{\sqrt{	an x}} - e^{\sqrt{x}}}{\sqrt{	an x} - \sqrt{x}}$. Let $u = \sqrt{	an x} - \sqrt{x}$. As $x ightarrow 0^{+}$,$u ightarrow 0$. Using the limit $\lim_{u ightarrow 0} \frac{e^{u+\sqrt{x}} - e^{\sqrt{x}}}{u} = e^{\sqrt{x}} \lim_{u ightarrow 0} \frac{e^u - 1}{u} = e^{\sqrt{x}} \cdot 1$. At $x=0$,$\alpha = e^0 = 1$. Next,evaluate $\beta$: $\beta = \lim_{x ightarrow 0} (1 + \sin x)^{\frac{1}{2} \cot x} = e^{\lim_{x ightarrow 0} \frac{1}{2} \cot x \cdot \sin x} = e^{\lim_{x ightarrow 0} \frac{1}{2} \cos x} = e^{1/2} = \sqrt{e}$. The quadratic equation is $ax^2 + bx - \sqrt{e} = 0$. Since $\alpha = 1$ and $\beta = \sqrt{e}$ are roots,the sum of roots $\alpha + \beta = 1 + \sqrt{e} = -b/a$ and the product of roots $\alpha \beta = 1 \cdot \sqrt{e} = -\sqrt{e}/a$. From the product,$\sqrt{e} = -\sqrt{e}/a \implies a = -1$. From the sum,$1 + \sqrt{e} = -b/(-1) = b$. Thus,$a = -1$ and $b = 1 + \sqrt{e}$. Then $a + b = -1 + 1 + \sqrt{e} = \sqrt{e}$. Finally,$12 \log_e(a + b) = 12 \log_e(\sqrt{e}) = 12 	imes \frac{1}{2} = 6$.

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