Let f(x) = lim_{y rightarrow infty} y(x^{1/y} | vedclass

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(D) Given,$f(x) = \lim_{y \rightarrow \infty} y(x^{1/y} - 1)$.First,we simplify $f(x)$:$f(x) = \lim_{y \rightarrow \infty} \frac{x^{1/y} - 1}{1/y}$.Le

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$2024$

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(D) Given,$f(x) = \lim_{y ightarrow \infty} y(x^{1/y} - 1)$.First,we simplify $f(x)$:$f(x) = \lim_{y ightarrow \infty} \frac{x^{1/y} - 1}{1/y}$.Let $t = 1/y$. As $y ightarrow \infty$,$t ightarrow 0^+$.$f(x) = \lim_{t ightarrow 0} \frac{x^t - 1}{t}$.Using the standard limit $\lim_{t ightarrow 0} \frac{a^t - 1}{t} = \ln(a)$,we get $f(x) = \ln(x)$.Now,substitute $f(x) = \ln(x)$ into the given equation $2022 f(\frac{1}{x}) + P f(x) = f(x^2)$:$2022 \ln(\frac{1}{x}) + P \ln(x) = \ln(x^2)$.Since $\ln(\frac{1}{x}) = -\ln(x)$ and $\ln(x^2) = 2 \ln(x)$:$2022(-\ln(x)) + P \ln(x) = 2 \ln(x)$.$-2022 \ln(x) + P \ln(x) = 2 \ln(x)$.Dividing by $\ln(x)$ (for $x 
eq 1$):$-2022 + P = 2$.$P = 2024$.

Let $f(x) = \lim_{y \rightarrow \infty} y(x^{1/y} - 1)$,and $2022 f(\frac{1}{x}) + P f(x) = f(x^2)$,then $P =$

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