Krdytkyu

I have been struggled with the following problem.

Suppose that $f$ is differentiable on $(0, infty)$. If we have $lim_{x rightarrow infty} f(x) = c_1$ and $lim_{x rightarrow infty} f'(x) = c_2$, where $c_1 c_2$ are two constants.

Can we get $c_2 = 0$？ If not could you please show me a counterexample?

It is true, and here's a hint: if $n in mathbb{N}$ and $lim_{x to infty} f(x)$ exists, show that there exists some $M_n$ such that
$$x, y > M_n implies |f(x) - f(y)| < frac{1}{n}.$$
As such, if $x > M_n$, we get $|f(x + 1) - f(x)| < frac{1}{n}$. Use the Mean Value Theorem to find a sequence $x_n$ such that $x_n to infty$, but $f'(x_n) to 0$.

Full Answer:

Fix $n in mathbb{N}$. Then, since $lim_{x to infty} f(x) = c_1$, there exists some $M_n$ such that
$$x > M_n implies |f(x) - c_1| < frac{1}{2n}.$$
Then,
begin{align*}
x > M_n &implies x + 1 > M_n \
&implies |f(x) - c_1| + |c_1 - f(x + 1)| < frac{1}{2n} + frac{1}{2n} \
&implies |f(x) - f(x + 1)| < frac{1}{n}.
end{align*}
Let $a_n > max{M_n, n}$. Then $|f(a_n + 1) - f(a_n)| < frac{1}{n}$. Applying the mean value theorem, we get some $x_n in (a_n, a_n + 1)$ such that
$$|f'(x_n)| = left|frac{f(a_n + 1) - f(a_n)}{a_n + 1 - a_n}right| = |f(a_n + 1) - f(a_n)| < frac{1}{n}.$$
Hence, by squeeze theorem, $f'(x_n) to 0$. Note also that $x_n > a_n > n$, so $x_n to infty$.

Now, if $f'(x)$ converges as $x to infty$, then $f'(x_n)$ must converge to this limit. As we showed, this limit is $0$, so $f'(x) to 0$.

Use mean value theorem to get $f(x+1)-f(x)=f'(c)$ for some $c$ with $x<c<x+1$ and take limit as $xtoinfty $. You get $c_1-c_1=c_2$.