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(C) Let $L = \lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n$. Taking the natural logarithm on both sides: $\ln L = \lim _{n \r

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$b^{1 / a}$

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(C) Let $L = \lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n$. Taking the natural logarithm on both sides: $\ln L = \lim _{n \rightarrow \infty} n \ln \left(1 + \frac{b^{1 / n} - 1}{a}\right)$. Using the limit formula $\lim _{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1$,let $x = \frac{b^{1 / n} - 1}{a}$. As $n \rightarrow \infty$,$x \rightarrow 0$. $\ln L = \lim _{n \rightarrow \infty} n \cdot \left(\frac{b^{1 / n} - 1}{a}\right) \cdot \frac{\ln(1 + x)}{x} = \lim _{n \rightarrow \infty} \frac{n}{a} (b^{1 / n} - 1)$. Let $t = 1/n$. As $n \rightarrow \infty$,$t \rightarrow 0$. $\ln L = \frac{1}{a} \lim _{t \rightarrow 0} \frac{b^t - 1}{t}$. Since $\lim _{t \rightarrow 0} \frac{b^t - 1}{t} = \ln b$,we have: $\ln L = \frac{1}{a} \ln b = \ln(b^{1 / a})$. Therefore,$L = b^{1 / a}$.

If $a > 0, b > 0$,then $\lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n =$

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