<s>[INST] Calculating $\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}$ Maybe someone could explain the following to me:
So I want to calculate $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}$$
It is pretty much obvious that $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}=\dfrac{1}{2}$$
The thing that I don't understand is that if we use the known limit
$\lim _{x\rightarrow 0}\left( 1+x\right) ^{\dfrac{1}{x}}=e$, then:
$$\begin{aligned}\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}
=\\
\lim _{x\rightarrow 0^{+}}\left( \dfrac{2+x-1}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}=\\
\lim _{x\rightarrow 0^{+}}\left( \dfrac{2+x}{2+x}-\dfrac{1}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}
=\\
\lim _{x\rightarrow 0^{+}}\left( 1+\dfrac{1}{-x-2}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}\end{aligned}$$
And from here we will eventually get $$\lim _{x\rightarrow 0^{+}}\left( \dfrac{1+x}{2+x}\right) ^{\dfrac{1-\sqrt{x}}{1-x}}=\dfrac{1}{\sqrt{e}}$$
What am I missing? Thanks!
 [/INST] You didn't correctly use the known limit though, because one cannot here.
To be able to use the result $\lim_{z \rightarrow 0^+}(1+z)^{\frac{1}{z}} = e$ to evaluate $\lim_{x \rightarrow 0^+} \left(1+\frac{1}{-x-2}\right)^{\frac{1-\sqrt{x}}{1-x}}$, you would need both [among other things]

*

*the equation $\lim_{x \rightarrow 0^+} \frac{1}{-x-2} = 0$ to be true, [because put informally $\frac{1}{-x-2}$ is playing the role of $z$]


*and also to be true the equation $\lim_{x \rightarrow 0^+} \frac{1-\sqrt{x}}{1-x} = \infty$.
Neither of these is true though. As $x$ goes to $0$, $\frac{1}{-x-2}$ goes to $-\frac{1}{2}$ instead of $0$,  and $\frac{1-\sqrt{x}}{1-x}$ goes to $1$, not $\infty$.
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