for a differentiable function $f$ in $(0,\infty)$ and $ 0 First thing that came to my mind is uniform continuity because the derivative is bounded, but how can it serves me here? Thank you.
for a differentiable function $f$ in $(0,\infty)$ and $ 0
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1Hint: how would you prove that a differentiable function on $(0,\infty)$ with bounded derivative is uniformly continuous on $(0,\infty)$? Note that the condition in your question $0
0$) *does not* imply that the derivative of $f'$ is bounded on $(0,\infty)$ (since $\frac{1}{x^2}$ "blows up" if $x$ is close to $0$). On the other hand, it *does imply* that the derivative of $f$ is bounded on $(1,\infty)$. In any case, the uniform continuity of $f$ on $(1,\infty)$ does not really help since $(n+1)^2-n^2=2n+1\to \infty$ as $n\to\infty$. – 2011-06-22 -
1So Right, Thank you so much. – 2011-06-22
4 Answers
For the Lagrange's mean value theorem, there exists $\xi$ such that $n^2 < \xi < (n+1)^2$ and $$f((n+1)^2) - f(n^2) = f'(\xi)(2n + 1) < \frac {2n + 1} {\xi^2} < \frac {2n + 1} {n^4}$$
We get the result thanks to the inequalities $$0\leq \int_{n^2}^{(n+1)^2}f'(t)dt=f((n+1)^2)-f(n^2) \leq \int_{n^2}^{(n+1)^2}\frac 1{x^2} dx = \frac{-1}x\mid_{n^2}^{(n+1)^2}=\frac 1{n^2}-\frac 1{(n+1)^2}.$$
The mean value theorem can serve you here.
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1Here's a simple example where $f'(x) \downarrow 0$ as $x \to \infty$ but $f((n + 1)^2 ) - f(n^2 ) = 1$ for all $n$. Let $f(x)=\sqrt{x}$, $x > 0$. Then $f'(x)=\frac{1}{{2\sqrt x }}$ and $f((n + 1)^2 ) - f(n^2 ) = (n+1) - n =1$. – 2011-06-22
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0Shai: If $\lim_{x\to \infty}f'(x)=0$ what could we say about requested limit now? Isn't it the same? – 2011-06-22
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0Nir: In the above example of $f(x)=\sqrt{x}$, $\lim _{x \to \infty } f'(x) = 0$ and $\lim _{n \to \infty } (f((n + 1)^2 ) - f(n^2 )) = \lim _{n \to \infty } 1 = 1$. This indicates that you cannot conclude much from the condition $\lim _{x \to \infty } f'(x) = 0$. – 2011-06-22
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0But by applying the squeeze theorem you get that $f'(x)$ tends to 0 anyway, so what's the difference? You say that without knowing the specific information that was given it's not able to answer this question like the way we did? – 2011-06-22
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0Given a differentiable function $f$ on $(0,\infty)$, by the mean value theorem $f((n + 1)^2 ) - f(n^2 ) = f'(\xi) (2n+1)$ for some point $\xi = \xi(n)$ between $n^2$ and $(n+1)^2$. Since $\lim _{n \to \infty } (2n + 1) = \infty$, it is clear that the condition $\lim _{x \to \infty } f'(x) = 0$ is not enough to imply $\lim _{n \to \infty } (f((n + 1)^2 ) - f(n^2 )) = 0$; hence, indeed, further information is required. – 2011-06-22
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1Note that in the case of $f(x)=\sqrt{x}$, where $f'(x) = \frac{1}{{2\sqrt x }}$, it holds $\frac{1}{{2\sqrt {(n+1)^2 } }} \leq f'(\xi) \leq \frac{1}{{2\sqrt {n^2 } }}$, that is $\frac{1}{{2(n+1)}} \leq f'(\xi) \leq \frac{1}{{2n}}$; hence $\frac{{2n + 1}}{{2(n + 1)}} \le f'(\xi )(2n + 1) \le \frac{{2n + 1}}{{2n}}$, and so, by the squeeze theorem, $\lim _{n \to \infty } f'(\xi )(2n + 1) = 1$. – 2011-06-22
The following steps lead to a solution:
(1) Note the Mean Value Theorem in this context:
If $f$ is a differentiable function on $(0,\infty)$, then for all $a,b\in (0,\infty)$, $a
$f(b)-f(a)=f'(c)(b-a).$
(2) Deduce that for all positive integers $n$, we have $f((n+1)^2)-f(n^2)=f'(c_n)((n+1)^2-n^2)$ for some real number $c_n$ such that $n^2 (3) Show that $(n+1)^2-n^2=2n+1$ and $\frac{1}{c_n}<\frac{1}{n^2}$ for all positive integers $n$. (4) Deduce that $\left|f((n+1)^2)-f(n^2)\right|=\left|f'(c_n)\right|\left|(2n+1)\right|<\frac{2n+1}{c_n^2}<\frac{2n+1}{n^4}$. (5) Finally, conclude that $\lim_{n\to\infty} \left[f((n+1)^2)-f(n^2)\right]=0$. I hope this helps!