If $\,\lim_{x\to 0} \Big(f\big({a\over x}+b\big) - {a\over x}\,f'\big({a\over x}+b\big)\Big)=c,\,$ find $\,\lim_{x\to\infty} f(x)$ 
Let $f$ be differentiable on $\mathbb R$ . If $\lim_{x\to 0}f\big({a\over x}+b\big)\ne 0$ and $$\lim_{x\to 0} \Big(f\big({a\over x}+b\big) - {a\over x}\,f'\big({a\over x}+b\big)\Big)=c,$$ find $\lim_{x\to\infty} f(x)$.  

I had been using L' Hopital's rule, but after I thought it was wrong.
 A: Set
$$
g(x)=\left\{\begin{array}{ccc}
x\,f\big(\frac{a}{x}+b\big) & \text{if} & x>0,\\
0 & \text{if} & x=0.
\end{array}\right.
$$
Clearly, $g$ is continuous in $[0,\infty)$, as $\lim_{x\to 0}f\big(\frac{a}{x}+b\big)$ exists.
Next
$$
g'(0)=\lim_{x\to 0}\frac{g(x)-g(0)}{x}=\lim_{x\to 0}\,f\Big(\frac{a}{x}+b\Big)=A.
$$
We need to find $A$.
Meanwhile, for $x>0$,
$$
g'(x)=f\Big(\frac{a}{x}+b\Big)-\frac{a}{x}\,f'\Big(\frac{a}{x}+b\Big).
$$
But as the limit
$$
\lim_{x\to 0^+}g'(x)=\lim_{x\to 0^+}\bigg(f\Big(\frac{a}{x}+b\Big)-\frac{a}{x}\,f'\Big(\frac{a}{x}+b\Big)\bigg)=c,
$$
exists, then it has to be equal to $g'(0)$. Thus $A=c$.
We use the following theorem:
If $f: [a,\infty)\to\mathbb R\,$ is continuous, and differentiable in $(a,\infty)$, and $\lim_{x\to a^+} f(x)$ exists, then $f$ is also differentiable at $a$, and $\,f'(a)=\lim_{x\to a^+}f'(x)$.
Note. We have not used that $\lim_{x\to 0}f\Big({a\over x}+b\Big)\ne 0$.
A: Apply l'Hopital to
$$
\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{x-b}{\frac{x-b}{f(x)}}\;.
$$
Then you get an expression that is fully in the given limits.

$$
…=\lim_{x\to\infty}\frac{1}{\frac{f(x)-(x-b)f'(x)}{f(x)^2}}=\frac{\Bigl(\lim_{x\to\infty}f(x)\Bigr)^2}{c}\;.
$$
Consequently, $\lim_{x\to\infty}f(x)=c$.
