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Let $\,f:\left[ {0,\infty } \right) \to \mathbb R$ be a continuously differentiable function such that $$ \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = L~~and~~~\mathop {\lim }\limits_{x \to \infty } f'\left( x \right) = K $$

Both limits exists and finite.

How can I prove that $K=0$?

My try:

Assuming $K\ne0$. W.L.O.G: $k\gt0$ .

If that's the case then then $\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = L$ doesn't exist because the derivative is positive around $K$ and that is not possible because $f(x)$ would keep increasing over $L+\epsilon$ for all $\epsilon$. How can I formalize that?

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marked as duplicate by Guy Fsone, José Carlos Santos, Community Jan 29 '18 at 16:15

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By mean value theorem, for each $x$ there exists $c_x\in (x,x+1)$ such that

$$f'(c_x)=f(x)-f(x+1)$$

Hence since $c_x\to\infty$ as $x\to\infty$ we get

$$\lim_{x\to \infty}f'(x)=\lim_{x\to \infty}f'(c_x) =\lim_{x\to \infty}[f(x)-f(x+1)] =0 $$

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  • $\begingroup$ could you please explain with more detail why $\lim_{x\to \infty} c_x = \infty \Rightarrow \lim_{x \to \infty} f´(x) = \lim_{x \to \infty} f´(c_x)$? $\endgroup$ – creepyrodent Aug 8 '18 at 15:41
  • $\begingroup$ @dude3221 set $X= c_x$ then done. since we know the limit exists by assumption $\endgroup$ – Guy Fsone Aug 9 '18 at 6:52
  • $\begingroup$ but $x \notin (x,x+1)$ $\endgroup$ – creepyrodent Aug 9 '18 at 13:19

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