Limit of derivative of a differentiable function. 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.
 A: 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 $\infty$.
A: 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.
