$f$ differentiable. Show there's $c\in(0,\infty)$ so that $f^{\prime}(c)=0$

Question

let $f : \mathbb{R} \longrightarrow \mathbb{R}$ be differentiable for all $x \in \mathbb{R}$, and let $L\in\mathbb{R}$ so that $f(0)=L$ and $\underset{x\rightarrow\infty}{\lim}f\left(x\right)=L$.
prove there's $c\in(0,\infty)$ so that $f^{\prime}(c)=0$. I get that I'm supposed to use Rolle's theorem, but I can't find 2 x values so $f(x_1)=f(x_2)$.

Answer 1

Take any $x_0>0$. If there is another $x_1\ne x_0$ with $f(x_0)=f(x_1)$ we are done by Rolle's theorem. If there is no such $x_1$ argue using the hypothesis $f(x)\overset{x\to\infty}{\rightarrow} L$ why $x_0$ has to be an extremum of $f$, hence $f'(x_0)=0$.
Edit: Wlog assume $f(x_0)>L$. If there is some $0<x<x_0$ with $f(x)>f(x_0)$ then by the intermediate value theorem there is an $x_1\in(0,x)$ with $f(x_1)=f(x_0)$ and we are done. Assume there is some $x>x_0$ with $f(x)>f(x_0)$. Since $f(x)\to L$ there has to be a $y>x$ with $f(y)<f(x_0)$. So again we get an $x_1\in(x,y)$ with $f(x_1)=f(x_0)$.

Answer 2

Let $t=\arctan x$ and $g(t)=f(\tan t)$. Define $$ g(\frac\pi2)=\lim_{x\to\infty}f(x)=L $$ Then $g(t)$ is defined in $[0,\frac\pi2]$ and satisfies the condition of Rolle's Theorem. Therefore there is $c\in(0,\frac\pi2)$ such that $$ g'(c_1)=0. $$ Note $$ g'(t)=f'(\tan t)\sec^2t $$ and hence $g'(c_1)=0$ becomes $$ f'(\tan c_1)= 0. $$ Let $c=\tan c_1\in(0,\infty)$ and then $f'(c)=0$.