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If $f(x)$ satisfies $97 f(x)+m f\left(\frac{1}{x}\right)=0$, where $f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right), x>0$, then the value of $m$ is
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Verified Answer
The correct answer is:
97
We have,
$$
\begin{aligned}
& f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right)=\lim _{m \rightarrow 0} \frac{x^m-1}{m}, \text { where } m=\frac{1}{n} \\
& \quad=\lim _{m \rightarrow 0} x^m \log x=\log x \\
& \text { Again, } 97 f(x)+m f\left(\frac{1}{x}\right)=0 \\
& \Rightarrow 97 \log x+m \log \left(\frac{1}{x}\right)=0 \Rightarrow 97 \log x-m \log x=0 \\
& \Rightarrow 97 \log x=m \log x \Rightarrow m=97
\end{aligned}
$$
$$
\begin{aligned}
& f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right)=\lim _{m \rightarrow 0} \frac{x^m-1}{m}, \text { where } m=\frac{1}{n} \\
& \quad=\lim _{m \rightarrow 0} x^m \log x=\log x \\
& \text { Again, } 97 f(x)+m f\left(\frac{1}{x}\right)=0 \\
& \Rightarrow 97 \log x+m \log \left(\frac{1}{x}\right)=0 \Rightarrow 97 \log x-m \log x=0 \\
& \Rightarrow 97 \log x=m \log x \Rightarrow m=97
\end{aligned}
$$
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