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 right 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.
$$
G_1(x) = (1+\sqrt{x}\;)^{\sqrt{x}}
\\
\log G_1(x) = \sqrt{x}\log(1+\sqrt{x}\;) = \sqrt{x}\log\left(\sqrt{x}\left(1+\frac{1}{\sqrt{x}}\right)\right)
\\
=\sqrt{x}\left(\log\sqrt{x}+\log\left(1+\frac{1}{\sqrt{x}}\right)\right)
=\sqrt{x}\left(\frac{1}{2}\log x+\frac{1}{\sqrt{x}}+O\left(\frac{1}{x}\right)\right)
\\
G_1(x) = e^{(1/2)\sqrt{x}\log x} e^1 (1+o(1))
\\
G_1(x) \sim e\;x^{(1/2)\sqrt{x}}
$$