Question

Let \[ f(x)=\lim_{y\to\infty} y\left(x^{1/y}-1\right), \] and \[ 2022\,f\left(\frac{1}{x}\right)+P\,f(x)=f(x^2), \] then \(P=\)

• A. \(2020\)
• B. \(2021\)
• C. \(2023\)
• D. \(2024\)

Hint:

Remember the important limit \[ \lim_{n\to\infty} n(a^{1/n}-1)=\log a. \] It follows directly from the expansion \[ a^{1/n}=e^{(\log a)/n}. \]

Answer 1

The Correct Option is D

Step 1: Evaluate \(f(x)\).
Given, \[ f(x)=\lim_{y\to\infty} y\left(x^{1/y}-1\right) \] Write \[ x^{1/y}=e^{\frac{\log x}{y}} \] Hence, \[ f(x) = \lim_{y\to\infty} y\left(e^{\frac{\log x}{y}}-1\right) \] Using the standard limit \[ \lim_{t\to 0}\frac{e^t-1}{t}=1, \] put \[ t=\frac{\log x}{y}. \] Then, \[ f(x) = \log x \cdot \lim_{t\to 0} \frac{e^t-1}{t} \] \[ f(x)=\log x. \]

Step 2: Find \(f\left(\frac1x\right)\) and \(f(x^2)\).
Since \[ f(x)=\log x, \] we have \[ f\left(\frac1x\right) = \log\left(\frac1x\right) = -\log x, \] and \[ f(x^2) = \log(x^2) = 2\log x. \]

Step 3: Substitute into the given relation.
Given, \[ 2022f\left(\frac1x\right)+Pf(x)=f(x^2). \] Substituting the values, \[ 2022(-\log x)+P(\log x)=2\log x. \] \[ (P-2022)\log x=2\log x. \]

Step 4: Compare coefficients.
Since \(\log x\neq 0\) in general, \[ P-2022=2. \] Therefore, \[ P=2024. \]

Step 5: Final conclusion.
Hence, \[ \boxed{2024} \]