# 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.

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?

• It seems correct.
– user439785
May 4, 2017 at 7:09
• It would have been nice to add the word "mean value theorem" somewhere in the proof, but apart from that, it looks great. May 4, 2017 at 7:53

## 1 Answer

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 $x\to \infty.$