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I wish to add a remark to Noam D. Elkies' beautiful answer. From the integral representation for $f$, putting $e^{-t}=s$ in the integral, $$f(-x)=1-x\int_0^\infty e^{-xte^{-t}} e^{-t}dt = 1-x\int_0^1 s^{sx}ds\ ,$$ so that, for $x\to \infty$, $ f(-x)=o(1)$ is equivalent to $$\int_0^1 xu(s)^xds=1+o(1)\ ,$$ where $u\in C([0,1])$ is the function $u(s):=s^s$. As a matter of fact, since $0\le u(s)\le 1$ for all $s$ and $u(s)=1$ only for $s=0$ or $s=1$, it turns out that the limit only depends on $u'(0)$ and $u'(1)$.

Since $u'(1)=1$, for any $\lambda < 1 < \mu$ there exists a $b < 1$ such that for all $s\in [b,1]$ there holds $$1+\mu(s-1) \le u(s)\le 1+\lambda(s-1)\ ,$$ so that $$x\big(1+\mu(s-1)\big)^x \le xu(s)^x\le x\big(1+\lambda(s-1)\big)^x\ .$$ Similarly, since $u'(0)=-\infty$, for any $\nu > 0$ there exists a $a > 0$ such that for all $s\in [0,a]$ $$u(s)\le1-\nu s\ ,$$ so $$xu(s)^x\le x\big( 1-\nu s\big) ^ x \ .$$

Moreover, since on any interval $[a,b]\subset\subset(0,1)$ the function $u$ is bounded away from $1$, it is clear that $\int_a^b xu(s)^xds=o(1)$ by uniform convergence to $0$.

Integrating over $s\in [ 0,1]$the inequalities above plainly give, and recalling that $\lambda < 1 < \mu$ and $\nu > 0$ were arbitrary, the inequalities above plainly give

$$\int_0^1 xu(s)^xds=\int_0^a x u(s)^xds+\int_a^b xu(s)^xds+\int_b^1 xu(s)^xds=1+o(1) \ ,$$ for $x\to \infty$.
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I wish to add a remark to Noam D. Elkies' beautiful answer. From the integral representation for $f$, putting $e^{-t}=s$ in the integral, $$f(-x)=1-x\int_0^\infty e^{-xte^{-t}} e^{-t}dt = 1-x\int_0^1 s^{sx}ds\ ,$$ so that, for $x\to \infty$, $ f(-x)=o(1)$ is equivalent to $$\int_0^1 xu(s)^xds=1+o(1)$$ (as $x\to xu(s)^xds=1+o(1)\ infty$), ,$$ where $u\in C([0,1])$ is the function $u(s):=s^s$. As a matter of fact, since $0\le u(s)\le 1$ for all $s$ and $u(s)=1$ only for $s=0$ or $s=1$, it turns out that the limit only depends on $u'(0)$ and $u'(1)$.

Since $u'(1)=1$, for any $\lambda < 1 < \mu$ there exists a $b < 1$ such that for all $s\in [b,1]$ there holds $$1+\lambda(s-1) $1+\mu(s-1) \le u(s)\le 1+\mu(s-1)1+\lambda(s-1)\ .$$ Similarly, since $u'(0)=-\infty$, for any $\nu > 0$ there exists a $a < > 0$ such that for all $s\in [0,\delta]$ 0,a]$ $$u(s)\le1-\nu s\ .$$

Moreover, since on any interval $[a,b]\subset\subset(0,1)$ the function $u$ is bounded away from $1$, it is clear that $\int_a^b xu(s)^xds=o(1)$ by uniform convergence to $0$.

Integrating over $s\in [ 0,1]$ the inequalities above plainly give, recalling that $\lambda < 1 < \mu$ and $\nu > 0$ were arbitrary,

$$\int_0^1 xu(s)^xds=\int_0^a x u(s)^xds+\int_a^b xu(s)^xds+\int_b^1 xu(s)^xds=1+o(1) \ ,$$ for $x\to \infty$.
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I wish to add a remark to Noam D. Elkies' beautiful answer. From the integral representation for $f$, putting $e^{-t}=s$ in the integral, $$f(-x)=1-x\int_0^\infty e^{-xte^{-t}} e^{-t}dt = 1-x\int_0^1 s^{sx}ds\ ,$$ so that $ f(-x)=o(1)$ is equivalent to $$\int_0^1 xu(s)^xds=1+o(1)$$ (as $x\to \infty$), where $u\in C([0,1])$ is the function $u(s):=s^s$. As a matter of fact, since $0\le u(s)\le 1$ for all $s$ and $u(s)=1$ only for $s=0$ or $s=1$, it turns out that the limit only depends on $u'(0)$ and $u'(1)$.

Since $u'(1)=1$, for any $\lambda < 1 < \mu$ there exists a $b < 1$ such that for all $s\in [b,1]$ there holds $$1+\lambda(s-1) \le u(s)\le 1+\mu(s-1)\ .$$ Similarly, since $u'(0)=-\infty$, for any $\nu > 0$ there exists a $a < 0$ such that for all $s\in [0,\delta]$ $$u(s)\le1-\nu s\ .$$

Moreover, since on any interval $[a,b]\subset\subset(0,1)$ the function $u$ is bounded away from $1$, it is clear that $\int_a^b xu(s)^xds=o(1)$ by uniform convergence to $0$.

Integrating over $s\in 0,1]$ the inequalities above plainly give, recalling that $\lambda < 1 < \mu$ and $\nu > 0$ were arbitrary,

$$\int_0^1 xu(s)^xds=\int_0^a x u(s)^xds+\int_a^b xu(s)^xds+\int_b^1 xu(s)^xds=1+o(1) \ ,$$ for $x\to \infty$.