Prove that $\lim\limits_{x\to+\infty}f'(x)=0$

Question

Give $f \in C^2(\mathbb{R})$. Prove that if $\lim\limits_{x\to+\infty}f(x)=a \in \mathbb{R}$ and $\lim\limits_{x\to+\infty}f''(x)=0$ then $\lim\limits_{x\to+\infty}f'(x)=0$.

My attempt: Get $x_0 > 0$. By using the Taylor formula to degree 2 at $x_0$ we have $$f(x)=f(x_0) + \frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2 + \mathcal{O}((x-x_0)^2).$$ Then, if $x=x_0+\frac{1}{x_0}$ we have $$f\left(x_0+\frac{1}{x_0}\right)=f(x_0)+\frac{f'(x_0)}{x_0} + \frac{f''(x_0)}{2x_0^2} + \mathcal{O}\left(\frac{1}{x_0^2}\right).$$ Let $x_0 \to +\infty$ we have $$\lim\limits_{x\to+\infty}\frac{f'(x_0)}{x_0}=0.$$ So there are two case: $\lim\limits_{x_0\to+\infty}f'(x_0)=0$ or $\lim\limits_{x_0\to+\infty}f'(x_0)=+\infty$ ( $f'(x_0)$ goes to $+\infty$ slower than $x_0$).

Question: How can I show $\lim\limits_{x_0\to+\infty}f'(x_0)=+\infty$ if my direction is correct? And if my direction is wrong, I hope you can hint the right way to me.

Answer 1

Try letting $x=x_0+1$ and using first degree Taylor polynomial with Lagrange remainder: $$f(x)=f(x_0)+f'(x_0)\cdot(x-x_0)+\dfrac{f''(c_{x,x_0})}{2}\cdot(x-x_0)^2, \ \ c_{x,x_0}\in(x,x_0)$$

Answer 2

Since $\lim\limits_{x\to+\infty}f(x)=a$ there is M such that for x > M we have $|f(x+1) - f(x)| < \epsilon$.
And from the mean value theorem, for every such x, there is y = y(x) between x and x+1 for which $f(x+1) - f(x) = f^{'}(y)$. In words, for every $\epsilon$ there is M such that for x > M in each interval [x,x+1] there is a point y where first derivative is less than $\epsilon$.
On the other hand, mean value theorem when applied to the first derivative gives $f^{'}(x+h) - f^{'}(x) = f^{''}(z)$ for some z between x and x+h.
Since second derivative tends to zero, we may assume that for x > M and any h > -1 we have $|f^{'}(x+h) - f^{'}(x)| < \epsilon$. If we now put for x value of y, where derivative is less than $\epsilon$ we see that $|f^{'}(y+h)| < 2\epsilon$ If you now let h very from -1 to 1 we get that derivative is no bigger than $2*\epsilon$ in the interval y-1, y+1. But we proved above that such y exists in every interval [x,x+1] for sufficiently large x. That means that derivative of f is less than $2*\epsilon$ for x > M, that is it tends to zero.