how to show $c=0$,$f:(0,\infty)\rightarrow\mathbb{R}$ is differentiable, $\lim_{x\rightarrow\infty}f(x)=1$, $f:(0,\infty)\rightarrow\mathbb{R}$ is differentiable, $\lim_{x\rightarrow\infty}f(x)=1$, $\lim_{x\rightarrow\infty}f'(x)=c$   we need to show $c=0$

well, I tried like this $|f(x)-1|&lt;\epsilon\forall x&gt;M$, where $M$ is very large, $|f'(x)-c|&lt;\epsilon\forall x&gt;M$,  what more I can say?thank you.
Since $\lim_{x \to \infty} \frac{f'(x)}{1}$ exists, by L'Hospital

$$\lim_{x \to \infty} \frac{f(x)}{x}=\lim_{x \to \infty} \frac{f'(x)}{1}=c \,.$$

But the first limit is $0$.

Alternately Use the MVT on the interval $[x,2x]$. You have

$$\frac{f(2x)-f(x)}{x}=f'(c_x)$$

Now, when $x \to \infty$ we have $f'(c_x) \to f(c)$ and the LHS gos to $0$.
If c isn't 0 then f is increasing/decreasing and can't converge to a limit.