Question:

Step 1: Recognize the infinite series as a geometric series. The given equation is: $$f'(x) + (f'(x))^2 + (f'(x))^3 + (f'(x))^4 + \cdots = e^x$$ Since $f'(x) \in (-1,1)$, this is a geometric series with first term $a = f'(x)$ and common ratio $r = f'(x)$. The series converges because $|f'(x)| < 1$. Step 2: Find the sum of the geometric series. For a geometric series with first term $a$ and common ratio $r$ where $|r| < 1$, the sum is $\frac{a}{1-r}$. Therefore: $$\frac{f'(x)}{1-f'(x)} = e^x$$ Step 3: Solve for $f'(x)$. Rearranging the equation from Step 2: $$f'(x) = e^x(1 - f'(x))$$ $$f'(x) = e^x - e^x \cdot f'(x)$$ $$f'(x) + e^x \cdot f'(x) = e^x$$ $$f'(x)(1 + e^x) = e^x$$ $$f'(x) = \frac{e^x}{1+e^x}$$ Step 4: Identify the form of the limit. We need to find: $$\lim_{x \to 0}(1+f(x))^{\frac{1}{x}}$$ Since $f(0) = 0$, as $x \to 0$, we have $1 + f(x) \to 1$. Also, the exponent $\frac{1}{x} \to \infty$. This is an indeterminate form of type $1^{\infty}$. Step 5: Apply the standard limit formula for indeterminate form $1^{\infty}$. For limits of the form $1^{\infty}$, we use: $$\lim_{x \to a} (1+g(x))^{h(x)} = \exp\left(\lim_{x \to a} g(x) \cdot h(x)\right)$$ In our case: $$\lim_{x \to 0}(1+f(x))^{\frac{1}{x}} = \exp\left(\lim_{x \to 0} f(x) \cdot \frac{1}{x}\right) = \exp\left(\lim_{x \to 0} \frac{f(x)}{x}\right)$$ Step 6: Evaluate the limit using L'Hôpital's Rule. Since $f(0) = 0$, the limit $\lim_{x \to 0} \frac{f(x)}{x}$ is of the form $\frac{0}{0}$. By L'Hôpital's Rule: $$\lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f'(x)}{1} = f'(0)$$ Step 7: Calculate $f'(0)$. Substituting $x = 0$ into the expression for $f'(x)$: $$f'(0) = \frac{e^0}{1+e^0} = \frac{1}{1+1} = \frac{1}{2}$$ Step 8: Find the final answer. Therefore: $$\lim_{x \to 0}(1+f(x))^{\frac{1}{x}} = \exp\left(\frac{1}{2}\right) = e^{1/2} = \sqrt{e}$$ The answer is **Option 2: $\sqrt{e}$**