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: Rewrite x^(1/y) using the exponential form.
x^(1/y) = e^((log x)/y). Then f(x) = lim_{y→∞} y[e^((log x)/y) - 1].

Step 2: Apply the standard exponential limit.
Let t = (log x)/y. As y → ∞, t → 0. Then f(x) = lim_{t→0} (log x)·(e^t - 1)/t = log x · 1 = log x.

Step 3: Compute f(1/x) and f(x²).
f(1/x) = log(1/x) = -log x. f(x²) = log(x²) = 2 log x.

Step 4: Substitute into the given functional equation.
2022(-log x) + P(log x) = 2 log x → (P - 2022) log x = 2 log x. Since log x ≠ 0 generally, P - 2022 = 2 → P = 2024.

Step 5: Final conclusion.
The value of P is 2024.