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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 $0$.

Bérénice
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user3615045
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  • Do you have an example of a function where the derivative is not 0? – Fabian May 06 '16 at 09:13
  • 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? – Fabian May 06 '16 at 09:15
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    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 – user5954246 May 06 '16 at 12:34

5 Answers5

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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 $0$.

Paramanand Singh
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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 $s=0$.

velut luna
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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.

Merkh
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  • The function is not a constant. I wonder how you got the idea that the function is constant. – Paramanand Singh May 06 '16 at 09:36
  • 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. – Merkh May 08 '16 at 21:51
  • Only when the function is constant on an interval the derivative is guaranteed to be zero on that interval. Otherwise not. – Paramanand Singh May 09 '16 at 06:28
  • What are you trying to say? Idon't see your point. Are you trying to say that $s \neq 0?$ – Merkh May 09 '16 at 10:37
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    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 $0$. – Paramanand Singh May 09 '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, $s=0$.

MathematicianByMistake
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  • 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. – Paramanand Singh May 06 '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, $s=0$.

Pratyush Yadav
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  • this is totally non-rigorous. In fact whatever symbolic manipulations you have does not make any sense. – Paramanand Singh May 06 '16 at 09:35
  • @Paramanand Can you at least tell me what is wrong with my method. I'd be happy to take my answer down if you can show me it is incorrect. – Pratyush Yadav May 06 '16 at 09:39
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    The equation starting from $f'(x) = \dfrac{f(\to \infty) - f(\to \infty)}{h}$ and next steps are completely wrong. – Paramanand Singh May 06 '16 at 09:41
  • @Paramanand Why? If $x+h \to \infty$ then $f(x+h) = 1$, because $\lim_{k \to \infty} f(k) = 1$. That makes the numerator exact zero. And that makes the derivative exact zero? I don't see what's wrong. – Pratyush Yadav May 06 '16 at 09:46
  • Also the asymptote thing you have written is wrong. If $f(x) \to 1$ as $x \to \infty$ then it does not mean that $y = 1$ is an asymptote (check $f(x) = 1 + (1/x)\sin(1/x)$ for example). – Paramanand Singh May 06 '16 at 09:47
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    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. – Paramanand Singh May 06 '16 at 09:48
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    -1) very poor usage of symbols – user5954246 May 06 '16 at 12:21
  • @ParamanandSingh - it appears you are interpreting "asymptote" to mean a line the function approaches without touching. You should be aware that the "no touching" part is not a universal convention. For many people (most in my experience), all that is necessary that the function approach the line. So $y = 1$ is an asymptote in this case. – Paul Sinclair May 06 '16 at 18:11
  • @PratyushYadav - to be more clear, the flaw in your proof (other than bad notation) is that you exchanged the two limits. Such exchanging of limits is not universally valid. It can only be justified in special cases (such as when the 2D limit exists). Since we know very little about $f$, it is impossible to justify it here. A simple example of when it fails$$\lim_{x\to 0+}\lim_{y\to 0+} x^y = 1\\lim_{y\to 0+}\lim_{x\to 0+} x^y = 0$$ – Paul Sinclair May 06 '16 at 18:28
  • @Paul thanks for clearing this up, especially the example. I'll keep this in mind. Also, apparently the notations we use in class aren't exactly popular. – Pratyush Yadav May 06 '16 at 18:36
  • Somewhat perhaps, but the real problem is that you used it poorly. In particular, you should have $f'(\to\infty)$ instead of $f'(x)$. – Paul Sinclair May 06 '16 at 18:37
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    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. – Paul Sinclair May 06 '16 at 18:41