I want to find a probability distribution (given by its CDF $F$) such that $$\liminf_{x \to \infty} \mathrm e^{\lambda x}(1-F(x)) = 0 \quad \text{for all } \lambda > 0$$ and $$\limsup_{x \to \infty} \mathrm e^{\lambda x}(1-F(x)) = \infty \quad \text{for all } \lambda > 0.$$
The only thing that I could guess almost immediately is that this distribituion has to be heavy-tailed. Otherwise I have no clue.
It suffices to find a non-negative integrable function $f$ such that for all $c>1$, there are sequences $(x_n'),(x_n'')$ tending to infinity such that you have $c^{x_n'}\int_{x_n'}^\infty f(x)\,\mathrm{d}x$ grows to infinity and $c^{x_n''}\int_{x_n''}^\infty f(x)\,\mathrm{d}x$ decays to zero.
The trick to deal with all $c$ at the same time is to make $\int_{x_n''}^\infty f(x)\,\mathrm{d}x$ decay superexponentially, while making $\int_{x_n'}^\infty f(x)\,\mathrm{d}x$ decay only subexponentially, for instance by ensuring that $\int_{x_n'}^\infty f(x)\,\mathrm{d}x\geq 1/(2x'_n)$, while $\int_{x_n''}^\infty f(x)\,\mathrm{d}x\leq 2^{-2^{x''_n}}$.
I'll leave the details to you, but with a clever choice of $(x_n')$ and $(x_n'')$, you can ensure that for $A=\bigcup_n (x_n',x_n'')$ (the union of intervals), this is true for the function $f(x)=1/x^2\cdot \mathbf 1_A(x)$ (where $\mathbf 1_A$ is the indicator function of $A$).
Then you only need to rescale $f$ to ensure that it is a pdf. $F$ can be recovered by integrating the resulting function, if needed.