Question

The value of the limit \[ \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{\frac{1}{x}} \] is:

• A. \( 1 \)
• B. \( 3^{1/3} \)
• C. \( 3^{1/4} \)
• D. \( 4^{1/4} \)

Hint:

For limits involving exponential terms and small \( x \), use approximations such as \( a^x \approx 1 + x \ln(a) \), and apply logarithms for simplification.

Answer 1

The Correct Option is D

To solve the given limit, we first analyze the expression: \[ \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{\frac{1}{x}}. \] For small values of \( x \), we can use the approximation \( a^x \approx 1 + x \ln(a) \) when \( x \) is close to 0. Applying this to each term: \[ 1^x \approx 1, \quad 2^x \approx 1 + x \ln(2), \quad 3^x \approx 1 + x \ln(3), \quad 4^x \approx 1 + x \ln(4). \] Thus, the numerator becomes: \[ 1 + (1 + x \ln(2)) + (1 + x \ln(3)) + (1 + x \ln(4)) = 4 + x(\ln(2) + \ln(3) + \ln(4)). \] Now the limit expression is: \[ \lim_{x \to 0} \left( \frac{4 + x (\ln(2) + \ln(3) + \ln(4))}{4} \right)^{\frac{1}{x}}. \] Simplifying the fraction: \[ = \lim_{x \to 0} \left( 1 + \frac{x (\ln(2) + \ln(3) + \ln(4))}{4} \right)^{\frac{1}{x}}. \] Using the approximation \( (1 + u)^n \approx e^{nu} \) for small \( u \), we get: \[ \lim_{x \to 0} \exp \left( \frac{(\ln(2) + \ln(3) + \ln(4))}{4} \right) = \exp \left( \frac{\ln(24)}{4} \right). \] This simplifies to: \[ \exp \left( \frac{\ln(24)}{4} \right) = 24^{1/4}. \] Thus, the correct answer is \( 4^{1/4} \).