Question:

Step 1: Recognize and sum the infinite geometric series. The given equation contains an infinite geometric series on the left side: $$f'(x) + (f'(x))^2 + (f'(x))^3 + (f'(x))^4 + \cdots$$ Since $f'(x) \in (-1, 1)$, this is a convergent geometric series with first term $f'(x)$ and common ratio $f'(x)$. Using the formula for an infinite geometric series: $$\sum_{n=1}^{\infty} (f'(x))^n = \frac{f'(x)}{1 - f'(x)}$$ Step 2: Set up and solve the equation for $f'(x)$. According to the problem, this sum equals $e^x$: $$\frac{f'(x)}{1 - f'(x)} = e^x$$ Multiply both sides by $(1 - f'(x))$: $$f'(x) = e^x(1 - f'(x))$$ Expand and rearrange: $$f'(x) = e^x - e^x \cdot f'(x)$$ $$f'(x) + e^x \cdot f'(x) = e^x$$ $$f'(x)(1 + e^x) = e^x$$ Therefore: $$f'(x) = \frac{e^x}{1 + e^x}$$ Step 3: Find $f'(0)$. Substitute $x = 0$ into the expression for $f'(x)$: $$f'(0) = \frac{e^0}{1 + e^0} = \frac{1}{1 + 1} = \frac{1}{2}$$ Step 4: Identify the form of the limit and apply logarithm. We need to evaluate: $$L = \lim_{x \to 0}(1 + f(x))^{\frac{1}{x}}$$ Since $f(0) = 0$, as $x \to 0$, we have $(1 + f(x)) \to 1$ and the exponent $\frac{1}{x} \to \infty$. This is an indeterminate form of type $1^{\infty}$. Taking the natural logarithm: $$\ln L = \lim_{x \to 0} \frac{1}{x} \ln(1 + f(x)) = \lim_{x \to 0} \frac{\ln(1 + f(x))}{x}$$ Step 5: Apply L'Hôpital's Rule. Since $f(0) = 0$, both the numerator and denominator approach $0$ as $x \to 0$, giving the indeterminate form $\frac{0}{0}$. Apply L'Hôpital's Rule: $$\ln L = \lim_{x \to 0} \frac{\frac{d}{dx}[\ln(1 + f(x))]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{\frac{f'(x)}{1 + f(x)}}{1} = \lim_{x \to 0} \frac{f'(x)}{1 + f(x)}$$ Step 6: Evaluate the limit and find $L$. Substitute $x = 0$: $$\ln L = \frac{f'(0)}{1 + f(0)} = \frac{\frac{1}{2}}{1 + 0} = \frac{1}{2}$$ Therefore: $$L = e^{\frac{1}{2}} = \sqrt{e}$$ The value of $\displaystyle\lim_{x \to 0}(1 + f(x))^{\frac{1}{x}} = \sqrt{e}$, which corresponds to **Option 2**.