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  1. Let $f\colon(0,\infty)\to\mathbb R$ be differentiable on $(0,\infty)$ and suppose that $\lim\limits_{x\to\infty} f'(x)=L$.

a. Show that for any $h>0$, $\lim\limits_{x\to\infty} \frac{f(x+h)-f(x)}h=L$.

b. Show that $\lim\limits_{x\to\infty} \frac{f(x)} {x} =L$.

this is a differentiation question.

i don't know how to solve this question

Because it is Korean, it is difficult to deal with this cite. So for me, it is not easy to write.

thank you

Paramanand Singh
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장원봉
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  • What are your thoughts? Is there anything your instructor has mentioned recently that would give you an inkling? – gen-ℤ ready to perish Dec 03 '17 at 08:10
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    @ Pippo a isn't completely the definition of the derivative – Netivolu Dec 03 '17 at 08:10
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    Related: https://math.stackexchange.com/questions/311152/applying-mvt-to-function-that-has-limit-of-its-derivative-given-when-x-to-infty, https://math.stackexchange.com/questions/1248374/let-f0-infty-to-r-be-differentiable-on-0-infty-and-fx-to-b-as, https://math.stackexchange.com/questions/1044668/if-fx-rightarrow-l-as-x-rightarrow-infty-infty-leq-l-leq-infty – Hans Lundmark Dec 03 '17 at 09:11

2 Answers2

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(a) comes from the Mean Value Theorem. For $\varepsilon>0$ there is $x_0$ such that $L-\varepsilon<f'(x)<L+\varepsilon$ for all $x\ge x_0$. If $x\ge x_0$ then $(f(x+h)-f(x))/h=f'(x+\xi)$ where $0<\xi<h$. Then $L-\varepsilon<f'(x+\xi)<L+\varepsilon$ etc.

Angina Seng
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For b). Given $\epsilon>0$, find some $M$ such that $|f'(x)-L|<\epsilon$ for all $x\geq M$. Now for all $x>M$, \begin{align*} \left|\dfrac{f(x)}{x}-L\right|&=\left|\dfrac{f(x)-f(M)}{x-M}\dfrac{x-M}{x}-L+\dfrac{f(M)}{x}\right|\\ &=\left|f'(\xi_{x})\cdot\dfrac{x-M}{x}-L+\dfrac{f(M)}{x}\right|\\ &=\left|(f'(\xi_{x})-L)\cdot\dfrac{x-M}{x}+\dfrac{x-M}{x}\cdot L-L+\dfrac{f(M)}{x}\right|\\ &\leq\epsilon\left|1-\dfrac{M}{x}\right|+|L|\cdot\dfrac{|M|}{x}+\dfrac{|f(M)|}{x}, \end{align*} taking $x\rightarrow\infty$ we have $\limsup_{x\rightarrow\infty}\left|\dfrac{f(x)}{x}-L\right|\leq\epsilon$, since $\epsilon>0$ is arbitrary, the result follows.

user284331
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