When $f(x)=-\ln(f(x))-x+1$, find $\lim_{x \to \infty} f(x)$ I want to find $\lim_{x \to \infty} f(x)$ when:
$$f(x)=-\ln(f(x))-x+1$$ 
Attempt:
I found that $f(0)=1$ and $f(1.19)\approx \frac{1}{2}$, so it seems that when $x$ grows, $f(x)$ goes to $0$. 
$f(x)$ cant be less than $0$ because this would break $$f(x)=-\ln(f(x))-x+1$$
So with this logic I believe $\lim_{x \to \infty} f(x)=0$. Is this a reasonable answer?
 A: Hint :
$$
\begin{align*}
& f(x) = - \ln f(x) - x + 1 \\
\iff & f(x) + \ln f(x) = 1-x \\
\iff & \ln e^{f(x)} + \ln f(x) = 1-x \\
\iff & \ln(e^{f(x)}f(x)) = 1-x \\
\iff & e^{f(x)}f(x) = e^{1-x} \\
\implies & \lim_{x \to \infty} e^{f(x)}f(x) = \lim_{x \to \infty} e^{1-x} \\
\iff & \lim_{x \to \infty} e^{f(x)}f(x) = 0.
\end{align*}
$$
A: We rewrite as $e^{1-x}=e^{f(x)}f(x)$. Taking limit $x\to\infty$ we get $\displaystyle\lim_{x\to\infty}e^{f(x)}f(x)=0$.
$$\displaystyle\lim_{x\to\infty}f(x)=0$$
$\Bigg($Though $e^{f(x)}$ grows (or decays) faster than $f(x)$, so $\displaystyle\lim_{x\to\infty}f(x)=-\infty$ but $\ln f(x)$ is undefined. $\Bigg)$
A: In the other answers it was shown that we get
$$ \lim_{x\rightarrow \infty} e^{f(x)} f(x) = 0. $$
Let me prove some other things here. First of all, that there exists a unique $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for all $x\in \mathbb{R}$ holds
$$ f(x) =- \ln(f(x)) -x +1.$$
To do so, we define
$$ g: \mathbb{R}_{>0} \rightarrow \mathbb{R},  \  g(y) = - \ln(y) - y +1.$$
Hence, the function $f$ is implicitely defined via the equation
$$ g(f(x)) = x.$$
Thus, all we need to prove is that $g$ is a bijection. Injectivity follows from
$$ g'(y) = - \frac{1}{y} - y < 0.$$
Surjectivity follows from the continuity of $g$ and
$$ \lim_{y\rightarrow \infty} g(y)= -\infty, \qquad \lim_{y\rightarrow 0} g(y)= \infty. $$ 
Let me now prove that indeed $\lim_{x\rightarrow \infty} f(x) = 0$ (in fact it was not clear to me in the first place why this limit should exist at all). First we pick a sequence $(x_n)_{n\geq 0}\subseteq \mathbb{R}$ such that $\lim_{n\rightarrow \infty} x_n = \infty $. The equality
$$ f(x) = - \ln(f(x)) -x +1 $$
tells us that $f(x_n)>0$ for all $n\geq 0$. Either the sequence $(f(x_n))_{n\geq 0}$ is unbounded or it is bounded.
Let us first assume that $(f(x_n))_{n\geq 0}$ is unbounded. As $\lim_{y \rightarrow \infty} e^y = \infty$ this violates $ \lim_{x\rightarrow \infty} e^{f(x)} f(x) = 0. $
Hence, we are left with the bounded case. Every subsequence $(f(x_{n_k}))_{k\geq 0}$ admits a converging subsequence $(f(x_{n_{k_l}}))_{l\geq 0}$. Assume that $f(x_{n_{k_l}}) \rightarrow \xi$. Then we get
$$0 = \lim_{l\rightarrow \infty} e^{f(x_{n_{k_l}})}f(x_{n_{k_l}}) = e^\xi \xi. $$
As $e^\xi >0$, we get $\xi =0$. We proved that every subsequence of $(f(x_n))_{n\geq 0}$ admits a converging subsequence and it converges to zero. Thus, we get
$$ \lim_{n\rightarrow \infty} f(x_n) = 0. $$
However, the sequence $(x_n)_{n\geq 0}$ with $\lim_{n\rightarrow \infty} x_n = \infty$ was arbitrary, and we therefore conclude
$$ \lim_{x\rightarrow \infty} f(x) = 0. $$
Of course one could do this proof using $g$, however, I wanted to remain close to previous answers.
