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$f(x)$ is a differentiable function on the real line such that $ \lim_{x\to \infty } f(x) =1 $ and $ \lim_{x\to \infty } f'(x) = s $ .Then

  1. $s$ should be $0$
  2. $s$ need not be $0$ but $|s| < 1$
  3. $s > 1$
  4. $s < -1$

Because $f(x)$ is bounded need not mean it can neither be increasing nor decreasing.So the derivative needs not be $0$.

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marked as duplicate by Hans Lundmark, user228113, Daniel W. Farlow, Martin Sleziak, Watson May 6 '16 at 18:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you have an example of a function where the derivative is not 0? $\endgroup$ – Fabian May 6 '16 at 9:13
  • $\begingroup$ You are given that $\lim_{x\to \infty} \lim_{h\to 0} (f(x+h)-f(x))/h =s$. Can you change the order of the limits? $\endgroup$ – Fabian May 6 '16 at 9:15
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    $\begingroup$ you can think in this way $f(x)=1+\frac{n}{\lambda}$, where $n \in \mathbb{N}$ and $\lambda =$ some polynomial. when limit tends to infinity $f(x)$ will tend to 1 and similarly derivative will tend to zero :D $\endgroup$ – user5954246 May 6 '16 at 12:34
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By Mean Value theorem we have the following equation $$f(x + 1) - f(x) = f'(\xi)$$ where $x < \xi < x + 1$. The LHS of the above equation tends to $1 - 1 = 0$ as $x \to \infty$ and RHS tends to $s$ as $x \to \infty$. Hence $s$ must be $0$.

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  • $\begingroup$ why the downvote? $\endgroup$ – Paramanand Singh May 7 '16 at 2:50
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Since $$\lim_{x\to\infty}f'(x)=s$$

That means for all $\epsilon>0$, there exists $R$ and $\delta>0$ s.t. $$\bigg|\frac{f(x+h)-f(x)}{h}-s\bigg|<\epsilon$$ whenever $x>R$ and $h<\delta$.

In other words, $$s-\epsilon<\frac{f(x+h)-f(x)}{h}<s+\epsilon$$

Now suppose $s\ne 0$, then set $\epsilon<|s|/2$ and $h=\delta/2$, we have $$\bigg|f(x+\delta/2)-f(x)\bigg|>\frac{|s|\delta}{4}$$

However, since $$\lim_{x\to\infty}f(x)=1$$ we know that there exist $R'$ so that $$\bigg|f(x)-1\bigg|<\frac{|s|\delta}{8}$$ whenever $x>R'$.

Now for $x>\max(R,R')$, we have $$\bigg|f(x+\delta/2)-f(x)\bigg| \le |f(x+\delta/2)-1|+|f(x)-1|<\frac{|s|\delta}{4}$$ which is a contradiction.

Hence $s=0$.

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If the derivative does not approach zero at infinity, the function value will continue to change (non-zero slope). Since we know the function is a constant, the derivative must go to zero.

Just pick an $|s| < 1,$ and draw what happens as you do down the real line. If $s \neq 0,$ the function can't remain a constant.

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  • $\begingroup$ The function is not a constant. I wonder how you got the idea that the function is constant. $\endgroup$ – Paramanand Singh May 6 '16 at 9:36
  • $\begingroup$ Since we know the function is a constant at infinity, the derivative must go to zero. Obviously the function isn't required to be constant over the whole line. $\endgroup$ – Merkh May 8 '16 at 21:51
  • $\begingroup$ Only when the function is constant on an interval the derivative is guaranteed to be zero on that interval. Otherwise not. $\endgroup$ – Paramanand Singh May 9 '16 at 6:28
  • $\begingroup$ What are you trying to say? Idon't see your point. Are you trying to say that $s \neq 0?$ $\endgroup$ – Merkh May 9 '16 at 10:37
  • $\begingroup$ I am saying that indeed $s=0$, but not because $f$ is constant. The real reason for $s=0$ is because of mean value theorem and the fact that $f'(x)$ tends to a limit as $x\to\infty$. See counter example $f(x)=1+\dfrac{\sin(x^{2})}{x}$. Here $f$ tends to $1$, but $f'$ does not tend to $0$. Limit of $f$ at infinity alone is not sufficient to conclude that $f'$ tends to $0$. To use your phrase, if $f$ is constant at infinity, it does not necessaroly mean that $f'$ tends to $0$. $\endgroup$ – Paramanand Singh May 9 '16 at 11:31
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Assume that the tangent of $f(x)$ is given by $$g(x)=mx+k$$ where m is the slope. But $m=\frac{dy}{dx}=f'(x)$

So $$g(x)=f'(x)x+k$$ As $x\rightarrow \infty$, we know that the asymptote is the horizontal line $y=1$, and it follows that $$\lim_{x\rightarrow \infty} g(x)=1\\\lim_{x\rightarrow \infty}(f'(x)x+k) =1$$

For this to be true, we must have an indeterminate form concerning the product $f'(x)x$. But $\lim_{x\rightarrow \infty} x=\infty$, so this can only happen if $\lim_{x\rightarrow \infty} f'(x)=0$

Thus, $s=0$.

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  • $\begingroup$ Again a totally non rigorous answer. The tangent $y=mx +k$ should be considered at some specific point. Otherwise if we chose $m=f'(x)$ then $k$ is not constant but rather a variable and your solution does not work. Also we have no information on asymptote. We just know that $f(x) \to 1$ as $x \to \infty$ and this does not guarantee asymptote. $\endgroup$ – Paramanand Singh May 6 '16 at 15:12
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Using the definition of derivative, $$f'(x)_{\text{at } x\to \infty} = \lim_{x \to \infty} \lim_{h\to 0+} \frac{f(x+h) - f(h)}{h} $$

But, since $x \to \infty$, $(x + h) \to \infty$

So, $$\begin{align} f'(x) &= \frac{f(\to \infty) - f (\to \infty)}{h} \\ &= \frac{1-1}{h} \\ &= \frac 0 h \\ &= 0 \end{align}$$

So, $s=0$.

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  • $\begingroup$ this is totally non-rigorous. In fact whatever symbolic manipulations you have does not make any sense. $\endgroup$ – Paramanand Singh May 6 '16 at 9:35
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    $\begingroup$ The equation starting from $f'(x) = \dfrac{f(\to \infty) - f(\to \infty)}{h}$ and next steps are completely wrong. $\endgroup$ – Paramanand Singh May 6 '16 at 9:41
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    $\begingroup$ the point is that you can not take limits $x \to \infty$ inside $\lim_{h \to 0}\dfrac{f(x + h) - f(x)}{h}$ You first need to evaluate the limit in $h$ and then take limit as $x \to \infty$. Moreover you have not used the assumption $f'(x) \to s$ as $x \to \infty$ which is crucial to the question. $\endgroup$ – Paramanand Singh May 6 '16 at 9:48
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    $\begingroup$ -1) very poor usage of symbols $\endgroup$ – user5954246 May 6 '16 at 12:21
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    $\begingroup$ I will say, though, that claiming your post made no sense was too strong. The exchange of limits is a fatal flaw in general, but for well-behaved functions, they are exchangeable. And the notation needs some correction, but the intent was clear. $\endgroup$ – Paul Sinclair May 6 '16 at 18:41

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