1. We need to evaluate

$$L=\lim_{x\to 0}\frac{e-(1+2x)^{\frac{1}{2x}}}{x}.$$

2. Let

$$y=(1+2x)^{\frac{1}{2x}}.$$

Then

$$\ln y=\frac{1}{2x}\ln(1+2x).$$

3. Expand $\ln(1+2x)$ near $x=0$:

$$\ln(1+2x)=2x-2x^2+O(x^3).$$

Therefore,

$$\frac{1}{2x}\ln(1+2x)=\frac{1}{2x}(2x-2x^2+O(x^3))=1-x+O(x^2).$$

So,

$$y=(1+2x)^{\frac{1}{2x}}=e^{1-x+O(x^2)}=e\,e^{-x+O(x^2)}.$$

Using $e^t=1+t+O(t^2)$ for small $t$,

$$e^{-x+O(x^2)}=1-x+O(x^2).$$

Hence,

$$(1+2x)^{\frac{1}{2x}}=e(1-x+O(x^2))=e-ex+O(x^2).$$

4. Substitute into the limit:

$$e-(1+2x)^{\frac{1}{2x}}=e-(e-ex+O(x^2))=ex+O(x^2).$$

Thus,

$$\frac{e-(1+2x)^{\frac{1}{2x}}}{x}=e+O(x).$$

Taking $x\to 0$,

$$L=e.$$

5. Therefore the correct option is

$$\boxed{D\;:\; e}.$$

6. Comparison with stored correct answer: stored answer is $D$, which matches our result.