Question:

Step 1: Identify the condition for continuity. For $f(x)$ to be continuous at $x = 0$, we need: $$k = \lim_{x \to 0} \left(\frac{2^x + 3^x + 5^x}{3}\right)^{3/x}$$ Step 2: Apply logarithm to simplify the limit. Taking the natural logarithm of the expression inside the limit: $$\ln k = \lim_{x \to 0} \frac{3}{x} \cdot \ln\left(\frac{2^x + 3^x + 5^x}{3}\right)$$ Step 3: Expand exponential functions using Taylor series. As $x \to 0$, we use the approximation $a^x \approx 1 + x\ln a$ for small $x$: $$2^x \approx 1 + x\ln 2$$ $$3^x \approx 1 + x\ln 3$$ $$5^x \approx 1 + x\ln 5$$ Step 4: Find the sum of the exponential terms. Adding these approximations: $$2^x + 3^x + 5^x \approx 3 + x(\ln 2 + \ln 3 + \ln 5) + O(x^2)$$ Step 5: Simplify the fraction inside the logarithm. Dividing by 3: $$\frac{2^x + 3^x + 5^x}{3} \approx 1 + \frac{x(\ln 2 + \ln 3 + \ln 5)}{3} + O(x^2)$$ Step 6: Apply logarithm to the simplified expression. Using $\ln(1 + u) \approx u$ for small $u$: $$\ln\left(\frac{2^x + 3^x + 5^x}{3}\right) \approx \frac{x(\ln 2 + \ln 3 + \ln 5)}{3}$$ Step 7: Evaluate the limit. Substituting back into the limit: $$\ln k = \lim_{x \to 0} \frac{3}{x} \cdot \frac{x(\ln 2 + \ln 3 + \ln 5)}{3}$$ $$\ln k = \lim_{x \to 0} (\ln 2 + \ln 3 + \ln 5)$$ $$\ln k = \ln 2 + \ln 3 + \ln 5 = \ln(2 \cdot 3 \cdot 5) = \ln 30$$ Step 8: Solve for $k$. Taking the exponential of both sides: $$k = e^{\ln 30} = 30$$ Therefore, the value of $k$ is **30**, which corresponds to **Option 4**.