Calculate limit involving exponents Calculate:
$$\lim_{x \rightarrow 0} \frac{(1+2x)^{\frac{1}{x}} - (1+x)^{\frac{2}{x}}}{x}$$
I've tried to calculate the limit of each term of the subtraction:
$$\lim_{x \rightarrow 0} \frac{(1+2x)^{\frac{1}{x}}}{x}$$
$$\lim_{x \rightarrow 0} \frac{(1+x)^{\frac{2}{x}}}{x} $$
Each of these two limits gave me $\lim_{x \rightarrow 0} e^{2 - \ln x}$, so the initial limit must be $0$. However, the correct result is $-e^2$ and I can't get it.
Please explain me what I did wrong and how to get the correct result.
Thank you!

Using Taylor expansion:
$$(1+2x)^{1/x}\approx e^2-2e^2x+o(x^2)$$
$$(1+x)^{2/x}\approx e^2-e^2x+o(x^2)$$
$$\lim_{x \rightarrow 0} \frac{(1+2x)^{\frac{1}{x}} - (1+x)^{\frac{2}{x}}}{x} \approx \lim _{x\to 0}\left(\frac{e^2-2e^2x+o\left(x^2\right)-e^2+e^2x+o\left(x^2\right)}{x}\right)$$
$$= \lim _{x\to 0}\left(\frac{-e^2x+o\left(x^2\right)}{x}\right)\rightarrow_0 \color{red}{-e^2}$$
