$$f:\mathbb R \rightarrow \mathbb R $$is a differentiable function. Given, $$\lim_{x\to \infty}f(x)=a$$ and $$\lim_{x\to \infty}f'(x)=b$$ how do you show that $b=0$?

I know that $f$ is a continuous function since it is differentiable, and it tends to $a$ at large x values. but how do i continue?

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    $\begingroup$ Think geometrically, what would it mean for $f$ if $f'$ tends to $b\neq 0$ at infinity? $\endgroup$ – Clément Guérin Mar 3 '16 at 8:32
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    $\begingroup$ $f(x+1)-f(x)=f'(\xi)$ $\endgroup$ – Hagen von Eitzen Mar 3 '16 at 8:34
  • $\begingroup$ Typo: $f(x)$, not $f(n)$. $\endgroup$ – Andrew Miloradovsky Mar 3 '16 at 8:41
  • $\begingroup$ @ClémentGuérin $f$ will be increasing or decreasing function which might not have a limit at large x? then what happens next? $\endgroup$ – winter Mar 3 '16 at 9:39

Since the function $f(x)$ is differentiable $g(x,h) = \frac{f(x+h)-f(x)}{h}$, is continuous on every point x, and h around 0 (zero).

$\lim_{x \to \infty} f'(x) = \lim_{x \to \infty} ( \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} )$

Because of the continuity interchange the limits and re-write the equation as

$\lim_{x \to \infty} f'(x) = \lim_{h \to 0} ( \lim_{x \to \infty} \frac{f(x+h)-f(x)}{h} )$

$\lim_{x \to \infty} f'(x) = \lim_{h \to 0} ( \frac{a-a}{h} ) = 0$

  • $\begingroup$ I don't understand your justification of interchange of the limits. For continuous (except one singular point $(X,Y)$) function $f(x,y)$ where $x,y\in\mathbb{R}$, $\lim_{x\rightarrow X}\lim_{y\rightarrow Y}f(x,y)$ and $\lim_{y\rightarrow Y}\lim_{x\rightarrow X}f(x,y)$ can be different. Consider, for example, $f(x,y)=(x^2-y^2)/(x^2+y^2)$. $\endgroup$ – Katie Imach Mar 3 '16 at 9:07
  • $\begingroup$ @KatieImach The key point in your example is $\lim_{(x,y) \to (0, 0)} f(x,y)$ does not exist. $\endgroup$ – crbah Mar 3 '16 at 9:13
  • $\begingroup$ so why does interchanging of limits work here? is there some sort of theorem? $\endgroup$ – winter Mar 3 '16 at 9:24
  • $\begingroup$ Ah, I see, existence of the limit $g(X)=\lim_{(x,h)\rightarrow (X,0)}\{(f(x+h)-f(x))/h\}$ itself is guaranteed by differentiability and continuity of $f(x)$ for $\forall X\in \mathbb{R}$ in this case (is my understanding correct?). But I'm still a bit confused about the case in $X=\infty$. Does the interchange work even if $g(X)\rightarrow \pm\infty$? If no, I think additional arguments to guarantee that $g(X)$ converges to a finite value is needed. $\endgroup$ – Katie Imach Mar 3 '16 at 9:31
  • $\begingroup$ @KatieImach No there is no need to set an extra condition. You can show this by playing epsilon-delta definitions of the limit. $\endgroup$ – crbah Mar 3 '16 at 9:45

If a derivative has a real limit $b$ in infinity, then (by the definition of the limit) for any positive $\epsilon$ there exists such (big enough) value $M$ that for $x$ above $M$ the derivative values $f'(x)$ differ from $b$ by less than $\epsilon$.

Let's see if $b$ can be non-zero. WLOG let's assume $b$ positive, then $\frac b2 < f'(x) < \frac 32b$ for $x$ being greater than some constant $T$. As $$f(x) = f(T) + \int\limits_{t=T}^x f'(t)\,dt$$ and $$f'(t)>\frac b2$$ above $T$, we have $$f(x)>f(T) + \frac b2 \cdot(x-T) = \text{some constant} + \frac b2 x$$ which means $f(x)$ grows unbound as $x$ grows to infinity. Which contradicts the assumption $f(x)$ has a finite limit of $a$.
Similar result for $b < 0$.

Finally $b$ must be zero for $f$ to have a finite limit.


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