$$\lim_{x \to 0} \frac{e - (1+2x)^{\frac{1}{2x}}}{x}$$ is equal to

Rewrite the term: Let $$f(x) = (1+2x)^{1/2x}$$. 

We know $$\lim_{x \to 0} f(x) = e$$.

Using the expansion $$a^b = e^{b \ln a}$$: $$(1+2x)^{1/2x} = e^{\frac{1}{2x} \ln(1+2x)}$$.

Series Expansion: $$\ln(1+2x) \approx 2x - \frac{(2x)^2}{2} = 2x - 2x^2$$.

So, $$\frac{1}{2x}(2x - 2x^2) = 1 - x$$.

$$f(x) \approx e^{1-x} = e \cdot e^{-x} \approx e(1 - x) = e - ex$$.

Substitute back:

$$\lim_{x \to 0} \frac{e - (e - ex)}{x} = \lim_{x \to 0} \frac{ex}{x} = \mathbf{e}$$.