Using the result \[ \lim_{x \to 0} (1 + kx)^{\frac{1}{x}} = e^k, \] the value of the function \( f(x) = (1 + 2x)^{\frac{1}{x}} \) as \( x \to 0 \) is: \[ e^2. \] Also, given that \( f(0) = e^2 \), to check the differentiability of the function, let: \[ y = (1 + 2x)^{\frac{1}{x}} \Rightarrow \log y = \frac{1}{x} \log(1 + 2x). \] \[ \Rightarrow \log y = \frac{1}{x} \left( 2x - \frac{4x^2}{2} + \ldots \right) = 2 - 2x + \ldots. \] Hence, \[ y = e^{2 - 2x + \ldots} = e^2 e^{-2x + \ldots} = e^2(1 - 2x + \ldots). \] Now, the right-hand derivative: \[ \frac{f(0+h) - f(0)}{h} = \frac{e^2(1 - 2h + \ldots) - e^2}{h} = -2e^2. \] Similarly, we can show that the left-hand derivative is also \( -2e^2 \). Therefore, the function is differentiable at \( x = 0 \).