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Let $f$ be a differentiable strictly decreasing positive function on $[0,\infty)$ then prove that $\lim_{x\to\infty} f^{'}(x)=0$. Using sequential criterion i proved that $\lim_{x\to\infty} f(x)$ exists and equal to inf $\{ f(x):x \in [0, \infty))\}$. Now i am confused first of all how to prove that
$\lim_{x\to\infty} f^{'}(x)$ exists ? Please help me about the existence of limit of the derivative. Graphically it seems that limit is zero but mathematically i want a proof of existence . Thanks for precise time.

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Using the generalized mean value theorem, we have $ \frac{\frac{f(2x)}{2x }-\frac{f(x)}{x}}{\frac{1}{2x}-\frac{1}{x}}=f(y)-yf^{'}(y)$ where $x<y<2x.$ Hence $\frac{f(2x)}{2x }-\frac{f(x)}{x}=\frac{y}{2x}(f^{'}(y)-\frac{f(y)}{y })$. This imply that $0 \leq|f^{'}(y)|\leq2|\frac{f(2x)}{2x }-\frac{f(x)}{x }|+|\frac{f(y)}{y }|$. Now since is $f$ is strictly decreasing $\frac{f(x)}{x }$ will tend to zero as x tends to infinity. Upon passage to the limit as $x\rightarrow \infty$ in last inequity we get the desired result.

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    $\begingroup$ If $y$ depends on $x$, how do you show that $y$ can actually take any positive value, so that $\lim_{y\to +\infty}f'(y)$ exists? $\endgroup$ – delt31 May 10 '17 at 11:46
  • $\begingroup$ Let's say, is there some $x$ guaranteed so that $\text{LHS}(x)=f(100)-100f'(100)$ where $x<100<2x$? $\endgroup$ – delt31 May 10 '17 at 11:59
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Actually the derivative need not tend to zero. You can construct a counter example in the following way:

For each natural number $n\ge 2$ the function $f_n:\left[-\frac{1}{n^2},\frac{1}{n^2}\right]\to \mathbb{R}$ given by

$f_n=\begin{cases} \frac1{n^2}-n^2\left(x+\frac1{n^2}\right)^2 && x<0\\ -\frac1{n^2}+n^2\left(x-\frac1{n^2}\right)^2 && x>0\\ \end{cases}$

is differentiable over its domain. It is strictly decreasing and satisfies $f'(0)=-2$. Also the difference between its maximum and minimum values is $\frac{2}{n^2}$. (The total "drop" in the function is $\frac{2}{n^2}$).

So start with a large $f(0)$, say $10$ will do. Define $f$ piecewise insert copies of $f_n$ around each natural number $n\ge 2$. If you want strictly decreasing $f$, insert linear pieces in between of extremely low negative derivative so that their "drop" is of the order of $\frac{1}{n^2}$. You will have to adjust the ends of the pieces accordingly.

The total "drop" in the function from $0$ to $\infty$ will then be dominated by the series $\sum \frac{1}{n^2}$ and thus will be finite. So you get a strictly decreasing positive function but the derivative at each natural number is $-2$ and therefore derivative does not approach $0$.

This is only a rough idea, I leave it to you to explicitly write down the formulae for each piece of the function.

Edit: Explicit construction for the function. The following function is, however, not strictly decreasing. Explicit formulae for the strictly decreasing case would be quite complicated.

Let $S_n=2\left(\dfrac1{2^2}+\dfrac1{3^2}+\cdots+\dfrac1{(n-1)^2}\right)+\dfrac1{n^2}$, $\left(S_2=\dfrac1{2^2}\right)$ and $K=10$ ($K$ could be anything greater than $\frac{\pi ^2}{3}$).

Define $f(x)=\begin{cases} K && 0\le x<2-\dfrac1{2^2}&&\\ K-S_n+f_n(x-n) && n-\dfrac1{n^2}\le x \le n+\dfrac1{n^2}&& n\ge 2\\ K-S_n-\dfrac1{n^2} && n+\dfrac1{n^2}< x < (n+1)-\dfrac1{(n+1)^2} && n\ge 2\\ \end{cases}$

Try drawing a rough graph of this function. It is easy to see that $f$ is (not strictly) decreasing, always positive (as $f(x)\ge K-\lim S_n= K-\frac{\pi ^2}{3}$) but $f'(n)=-2\quad \forall n\in \mathbb{N}$ and so $f'$ does not approach $0$ as $x$ approaches $\infty$.

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  • $\begingroup$ according to you above question is wrong?? $\endgroup$ – neelkanth Jun 12 '15 at 12:41
  • $\begingroup$ @YOGESH Yes, I am saying that the statement is false. However, in the example I wish to construct, the derivative does not approach any real number. It sort of oscillates between $-2$ and $0$ and stays near $-2$ for a very small interval, so that the "drop" in $f$ is very small. $\endgroup$ – Chaitanya Tappu Jun 12 '15 at 12:47
  • $\begingroup$ @YOGESH No, I construct a function such that limit of derivative does not exist. The derivative oscillates between $-2$ and $0$ but is always negative, so the function itself does not oscillate, but decreases strictly. $\endgroup$ – Chaitanya Tappu Jun 12 '15 at 12:51
  • $\begingroup$ your sequence of functions <fn(x)> is not convergent.... can you give an exact counterexample?? $\endgroup$ – neelkanth Jun 12 '15 at 12:55
  • $\begingroup$ if the limit of function as x tends to infinity exists say to b then line y=b will be asymptote to the graph of the function so limit of derivative goes to zero?? $\endgroup$ – neelkanth Jun 12 '15 at 12:58

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