Let's suppose $x/f(x) \to 0$. Then take logarithm. As $x \to \infty$, $$ \log G(x) = f(x) \log\left(1+\frac{x}{f(x)}\right) =f(x)\left(\frac{x}{f(x)}+O\left(\left(\frac{x}{f(x)}\right)^2\right)\right) \\ = x + O\left(\frac{x^2}{f(x)}\right) \tag{*}$$ Now if we have the stronger $x^2/f(x) \to 0$, then $$ G(x) = \exp\left(x + O\left(\frac{x^2}{f(x)}\right)\right) =e^x\exp \left(O\left(\frac{x^2}{f(x)}\right)\right) =e^x\left(1+o(1)\right) \tag{**}$$ But note: it is incorrect to rightwrite this as $G(x) \to e^x$. Instead write $G(x) \sim e^x$.
Now consider the three examples. See if I made any mistakes.
..........
$$
G_1(x) = (1+\sqrt{x}\;)^{\sqrt{x}}
\\
\log G_1(x) = \sqrt{x}\log(1+\sqrt{x}\;) = \sqrt{x}\log\left(\sqrt{x}\left(1+\frac{1}{\sqrt{x}}\right)\right)
\\
=\sqrt{x}\left(\log\sqrt{x}+\log\left(1+\frac{1}{\sqrt{x}}\right)\right)
=\sqrt{x}\left(\frac{1}{2}\log x+\frac{1}{\sqrt{x}}+O\left(\frac{1}{x}\right)\right)
\\
G_1(x) = e^{(1/2)\sqrt{x}\log x} e^1 (1+o(1))
\\
G_1(x) \sim e\;x^{(1/2)\sqrt{x}}
$$
..........
Assume $\alpha>1/2$
$$
G_2(x) = (1+x^\alpha)^{x^{1-\alpha}}
\\
\log G_2(x) = x^{1-\alpha}\log\left(1+x^\alpha\right)
=x^{1-\alpha}\left(\log x^\alpha+ \log\left(1+x^{-\alpha}\right)\right)
\\
=x^{1-\alpha}\left(\alpha \log x+x^{-\alpha} +O\left(x^{-2\alpha}\right)\right)
\\
G_2(x) = e^{x^{1-\alpha}\alpha\log x} e^{x^{1-2\alpha}} (1+o(1))
\sim x^{x^{1-\alpha}\alpha}
$$
..........
$$
G_3(x) = \left(1+\frac{x}{\log x}\right)^{\log x}
\\
\log G_3(x) = \log x \log\left(1+\frac{x}{\log x}\right)
=\log x\left(\log\frac{x}{\log x}+O\left(\frac{\log x}{x}\right)\right)
\\
=(\log x)^2 -\log x \log\log x+o(1)
\\
G_3(x) \sim e^{(\log x)^2-\log x \log\log x} = x^{\log x - \log\log x}
$$