What Didier is suggesting the use of the little oh notation. This stands for the following:
$e^x=1+x+o(x) \text{ for } x\to 0 $ means that
$\lim\limits_{x \to 0} \frac{e^x-1-x}{x} =0$
or that we can approximate $e^x$ near $x=0$ very accurately by $1+x$, with an error that is very small in comparison to $x$ i.e $\dfrac{o(x)}{x}\to0$ (that is what $o(x)$ kinda means.)
To make things easier, put $\sqrt n=x$, to get
$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{e^x} - {2^x}}}{x}} \right)^3}$
Now, we get the following:
$\eqalign{ & {e^x} = 1 + x + o\left( x \right) \cr & {2^x} = {e^{x\log 2}} = 1 + x\log 2 + o\left( x \right) \cr} $
So this means:
$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + x + o\left( x \right) - 1 - x\log 2 - o\left( x \right)}}{x}} \right)^3}$
$\mathop {\lim }\limits_{x \to 0} {\left( {1 - \log 2 + \frac{{o\left( x \right)}}{x}} \right)^3}$
But since we said $o(x)$ is an error that tends to zero in comparison to $x$, we get
$\mathop {\lim }\limits_{x \to 0} {\left( {1 - \log 2 + \frac{{o\left( x \right)}}{x}} \right)^3} = {\left( {1 - \log 2} \right)^3}$
Do you follow? Can you move on to the second case analogously?