If f(x) satisfies 97 f(x) + m fleft(frac{1}{x | vedclass

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(B) Given $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$. Let $h = \frac{1}{n}$. As $n \rightarrow \infty$,$h \rightarrow 0$. $f(x) = \lim_{h \ri

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$97$

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(B) Given $f(x) = \lim_{n ightarrow \infty} n(x^{1/n} - 1)$. Let $h = \frac{1}{n}$. As $n ightarrow \infty$,$h ightarrow 0$. $f(x) = \lim_{h ightarrow 0} \frac{x^h - 1}{h}$. Using the standard limit formula $\lim_{h ightarrow 0} \frac{a^h - 1}{h} = \ln a$,we get $f(x) = \ln x$. Now,substitute $f(x) = \ln x$ into the given equation: $97 f(x) + m f\left(\frac{1}{x}ight) = 0$ $97 \ln x + m \ln\left(\frac{1}{x}ight) = 0$ Since $\ln\left(\frac{1}{x}ight) = -\ln x$,we have: $97 \ln x - m \ln x = 0$ $(97 - m) \ln x = 0$ For this to hold for all $x > 0$ (where $\ln x 
eq 0$),we must have $97 - m = 0$,which implies $m = 97$.

If $f(x)$ satisfies $97 f(x) + m f\left(\frac{1}{x}\right) = 0$,where $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$ for $x > 0$,then the value of $m$ is:

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