Simple way to compute $\lim_{x \to 0}\left(\frac{1+x}{1+2x} \right)^{1/x}$ without de L'Hopital and Taylor

Question

What is the simplest way you can think of to compute the following limit?

$$\lim_{x \to 0}\left(\frac{1+x}{1+2x} \right)^{1/x}$$

Clearly, I can do it by using de L'Hopital or Taylor's theorems (and the properties of $\exp$ and $\log$), which give $$\lim_{x \to 0}\left(\frac{1+x}{1+2x} \right)^{1/x} = 1/e.$$

But I'm looking for a simpler way to do it by exploiting the fact that $$\lim_{x \to 0} (1+x)^{1/x} = e.$$

Answer 1

$$\left(\dfrac{1+x}{1+2x}\right)^{1/x}=\dfrac{(1+x)^{1/x}}{\left((1+2x)^{1/2x}\right)^2}.$$

Answer 2

Define $y:=1/x$

Then

$$\lim_{y \rightarrow \infty} \dfrac{(1+1/y)^y}{(1+2/y)^y}=\dfrac{\lim_{y \rightarrow \infty}(1+1/y)^y}{\lim_{y \rightarrow \infty}(1+2/y)^y}=e/e^2=e^{-1}$$

Where we used: $\lim_{y \rightarrow \infty}(1+a/y)^y =e^a$, for $a\in\Bbb R$.