TITLE: Prob. 5, Chap. 5 in Baby Rudin: If $f^\prime(x) \to 0$ as $x \to +\infty$, then $f(x+1) - f(x) \to 0$ as well.
QUESTION [5 upvotes]: Here is Prob. 5, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

Suppose $f$ is defined and differentiable for every $x > 0$, and $f^\prime(x) \to 0$ as $x \to +\infty$. Put $g(x) = f(x+1)-f(x)$. Prove that $g(x) \to 0$ as $x \to +\infty$. 

Here is an attempt of mine. 

As $$\lim_{x \to +\infty} f^\prime(x) = 0,$$ 
  so, given any real number $\varepsilon > 0$, we can find a real number $r$ such that $$\left\vert f^\prime(x) - 0 \right\vert < \varepsilon  \tag{1} $$
  for all real numbers $x$ which satisfy $x > r$. 
Let  $x$ be a real number such that $x > r$. Then as $f$ is continuous on $[ x, x+1]$ and differentiable on $(x, x+1)$, so by the Mean Value Theorem there is some point $p \in (x, x+1)$ for which 
  $$g(x) = f(x+1) - f(x) = \left( \ (x+1) - x \ \right) f^\prime(p) = f^\prime(p), $$ 
  and as $p > x > r$, so by (1) we can conclude that 
  $$\left\vert g(x) \right\vert = \left\vert f^\prime(p) \right\vert < \varepsilon.$$ 
  Thus, given a real number $\varepsilon > 0$, we can find a real number $r$ such that 
  $$\left\vert g(x) - 0 \right\vert < \varepsilon$$ 
  for all real numbers $x$ which satisfy $x > r$. 
Hence $$\lim_{x \to +\infty} g(x) = 0.$$

Is this proof correct?

REPLY [2 votes]: Yes, it's correct, and indeed the key is mean value theorem.  An intuitive way to see it is this:  As you had it, $g(x)=f(x+1)-f(x)=f'(p)(x+1-x)=f'(p)$ for some $p\in(x,x+1)$.  Now, as $x$ tends to infinity, clearly so does $p$.  But  $f'(p)\to 0$, as $p\to \infty.$  Hence $g(x)\to 0$, as $x\to \infty.$