0 like 0 dislike
72 views
Suppose $f$ is defined and differentiable for every $x>0$, and $f^{\prime}(x) \rightarrow 0$ as $x \rightarrow+\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x) \rightarrow 0$ as $x \rightarrow+\infty$.
| 72 views

0 like 0 dislike
Let $\varepsilon>0$. Choose $x_{0}$ such that $\left|f^{\prime}(x)\right|<\varepsilon$ if $x>x_{0}$. Then for any $x \geq x_{0}$ there exists $x_{1} \in(x, x+1)$ such that
$f(x+1)-f(x)=f^{\prime}\left(x_{1}\right) .$
Since $\left|f^{\prime}\left(x_{1}\right)\right|<\varepsilon$, it follows that $|f(x+1)-f(x)|<\varepsilon$, as required.
by Diamond (66,769 points)

1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike