Exercise 5.5

Question

Exercise 5: Suppose $f$ is defined and differentiable for every $x>0$, and $f'(x)\rightarrow0$ as $x\rightarrow+\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x)\rightarrow0$
    as $x\rightarrow+\infty$.

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon} ( extit{Matt ``Frito'' Lundy}) \ Fix $\varepsilon > 0$. $f'(x) o 0$ as $x o + \infty$ means that there exists an $M \in \mathbf R$ such that $x \geq M$ implies $\left| f'(x) ight| < \varepsilon$. Now for any $a, b \in \mathbf R$ such that $M < a < b$, ``the'' mean value theorem implies that there exists an $x \in (a,b)$ such that $$\frac{f(b) - f(a)}{b-a} = f'(x). $$ But $x \in (a,b)$ implies that $x > M$ so: $$\left| \frac{f(b) - f(a)}{b-a} ight| = \left| f'(x) ight| < \varepsilon.$$ Taking $b = a+1$ means if $a > M$, then $\left| g(a) ight| < \epsilon$, which shows that $g(x) o 0$ as $x o +\infty$.

Exercise 5.5

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Exercise 5.5

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