Step-by-step Solution:

\[ \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{1/x}. \] Step 1: Simplify the numerator Each term \(a^x\) (where \(a > 0\)) in the sum can be expressed as: \[ a^x = e^{x \ln a}. \] So, the numerator becomes: \[ 1^x + 2^x + 3^x + 4^x = e^{x \ln 1} + e^{x \ln 2} + e^{x \ln 3} + e^{x \ln 4}. \] Here, \(e^{x \ln 1} = 1\), as \(\ln 1 = 0\). For small \(x\), using the approximation \(e^{x} \approx 1 + x\), we expand: \[ e^{x \ln a} \approx 1 + x \ln a. \] Thus, the numerator becomes: \[ 1^x + 2^x + 3^x + 4^x \approx 1 + (1 + x \ln 2) + (1 + x \ln 3) + (1 + x \ln 4) = 4 + x (\ln 2 + \ln 3 + \ln 4). \] Step 2: Simplify the entire expression Divide the numerator by 4: \[ \frac{1^x + 2^x + 3^x + 4^x}{4} \approx 1 + \frac{x}{4} (\ln 2 + \ln 3 + \ln 4). \] Take the natural logarithm of the limit: \[ \ln \left[ \left( 1 + \frac{x}{4} (\ln 2 + \ln 3 + \ln 4) \right)^{1/x} \right]. \] Using the logarithmic property \(\ln (a^b) = b \ln a\), this becomes: \[ \frac{1}{x} \ln \left( 1 + \frac{x}{4} (\ln 2 + \ln 3 + \ln 4) \right). \] Step 3: Expand \(\ln(1 + u)\) For small \(u\), \(\ln(1 + u) \approx u\). Here, \(u = \frac{x}{4} (\ln 2 + \ln 3 + \ln 4)\). Thus: \[ \ln \left( 1 + \frac{x}{4} (\ln 2 + \ln 3 + \ln 4) \right) \approx \frac{x}{4} (\ln 2 + \ln 3 + \ln 4). \] Substituting back: \[ \frac{1}{x} \ln \left( 1 + \frac{x}{4} (\ln 2 + \ln 3 + \ln 4) \right) \approx \frac{1}{x} \cdot \frac{x}{4} (\ln 2 + \ln 3 + \ln 4). \] This simplifies to: \[ \frac{\ln 2 + \ln 3 + \ln 4}{4}. \] Step 4: Exponentiate to obtain the final result Exponentiating the result gives: \[ \exp \left( \frac{\ln 2 + \ln 3 + \ln 4}{4} \right) = \left( 2 \cdot 3 \cdot 4 \right)^{1/4}. \] Thus, the value of the limit is: \[ 4!^{1/4}. \]