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(A) Let $L = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{{a^{1/x}} + b}}{c}} ight)^x} = d$. Since $d$ is a non-zero finite value,the limi

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(A) Let $L = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{{a^{1/x}} + b}}{c}} ight)^x} = d$. Since $d$ is a non-zero finite value,the limit must be of the form $1^\infty$. As $x 	o \infty$,$a^{1/x} 	o a^0 = 1$. Thus,$\frac{1+b}{c} = 1$,which implies $c = b + 1$. Now,$L = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{{a^{1/x}} + b}}{b+1}} ight)^x} = \mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{{{a^{1/x}} - 1}}{{b+1}}} ight)^x}$. Using the standard limit $\mathop {\lim }\limits_{u 	o 0} (1+u)^{1/u} = e$,we have: $L = \exp \left( \mathop {\lim }\limits_{x 	o \infty } x \cdot \frac{a^{1/x} - 1}{b+1} ight)$. Since $\mathop {\lim }\limits_{x 	o \infty } x(a^{1/x} - 1) = \mathop {\lim }\limits_{t 	o 0} \frac{a^t - 1}{t} = \ln a$,we get: $d = \exp \left( \frac{\ln a}{b+1} ight) = a^{1/(b+1)}$. Taking $\log_a$ on both sides: $\log_a d = \frac{1}{b+1}$. Therefore,$(b + 1) \log_a d = 1$.

If $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{a^{1/x}} + b}}{c}} \right)^x} = d$ (where $d$ is a non-zero finite value),then $(b + 1) \log_a d$ is equal to:

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