1. ## differentiability and continuity

(a) Suppose that $f: \Re\rightarrow\Re$is differentiable and, that $f(x)/x\rightarrow 0$ as $x\rightarrow\infty$. Prove that, if $f'(x)\rightarrow k$ as $x\rightarrow\infty$, then $k=0$.
(b) Give an example of a function $g:\Re\rightarrow\Re$ such that $g(x)\rughtarrow 0$ as $x\rightarrow\infty$, but $g'(x)$ does not tend to a limit as $x\rightarrow\infty$

(a) I know f(x) must increase slower that x. So for some $N\in\Re$, $x_1,x_2\in\Re$, $|f(x_1)-f(x_2)|<|x_1-x_2|$. Now the trouble of showing that this inequality increases until $\frac{|f(x_1)-f(x_2)|}{x_1-x_2|}<\epsilon$. Dunno what to do!
(b) I tried all the functions I could think of. I figured it should be oscillating since we dont want the derivative to converge... But it doesnt seem to make since since a converged function/series isnt increasing or decreasing anymore. I thought of something like $(-1)^x/x$ but the derivative is complex. Otherwise I was trying things like sin(1/x) but couldnt find anything that worked.

2. a) it is not limitative to suppose that f(*) is differentiable in 0. In...

http://www.mathhelpforum.com/math-he...tml#post582679

... it has been extablished that if f(*) is differentiable in 0 is...

$\displaystyle f(x)= f(0) + k\ x + x\ \varepsilon (x) , \lim_{x \rightarrow 0} \varepsilon (x) =0$ (1)

From (1) we derive...

$\displaystyle f^{'}(x)= k + \varepsilon (x) + x\ \varepsilon^{'} (x)$ (2)

... and...

$\displaystyle \frac{f(x)}{x}= k + \varepsilon (x)$ (3)

... so that if $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = k$ for (3) is $\displaystyle \lim_{x \rightarrow \infty} \varepsilon(x)=0$ , if $\displaystyle \lim_{x \rightarrow \infty} f^{'} (x)= k$ for (2) is $\displaystyle \lim_{x \rightarrow \infty} x\ \varepsilon^{'} (x)=0$ and [finally...] if $\displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0$ that means that is $k=0$...

Kind regards

$\chi$ $\sigma$

3. any help on part b) anybody?

4. If the Laplace Tranform $\mathcal {L} \{\varphi(t)\}= f(s)$ exists , then the 'Final value theorem' extablishes that...

$\displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} s\ f(s)$ (1)

Now is...

$\displaystyle \mathcal{L} \{\varphi^{'} (t)\} = s\ f(s) - \varphi(0)$ (2)

... and taking into account (1) and the fact that is $\displaystyle \lim_{t \rightarrow \infty} \varphi(t)=0$ is...

$\displaystyle \lim_{t \rightarrow \infty} \varphi^{'}(t) = \lim_{s \rightarrow 0} \{ s^{2}\ f(s) - s\ \varphi(0) \} =0$ (3)

Kind regards

$\chi$ $\sigma$

5. Okay, $f(x)=\frac{\sin(x^3)}{x}$ works.

6. Originally Posted by chisigma
If the Laplace Tranform $\mathcal {L} \{\varphi(t)\}= f(s)$ exists , then the 'Final value theorem' extablishes that...

$\displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} s\ f(s)$ (1)

Now is...

$\displaystyle \mathcal{L} \{\varphi^{'} (t)\} = s\ f(s) - \varphi(0)$ (2)

... and taking into account (1) and the fact that is $\displaystyle \lim_{t \rightarrow \infty} \varphi(t)=0$ is...

$\displaystyle \lim_{t \rightarrow \infty} \varphi^{'}(t) = \lim_{s \rightarrow 0} \{ s^{2}\ f(s) - s\ \varphi(0) \} =0$ (3)

Kind regards

$\chi$ $\sigma$
In Gluskin’s paper ‘Let us teach this generalization of the final-value theorem’ [http://iopscience.iop.org/0143-0807/24/6/005/] there are some important details that usually You don’t find in ordinary complex analysis books. Among others in cap. 4 is written…

If $\displaystyle \lim_{t \rightarrow \infty} f(t)$ and the Laplace Transform $F(s) = \mathcal{L} \{f(t)\}$ exist, then we have the ‘final value theorem…

$\displaystyle \lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} s\ F(s)$ (4)

For this equality to be true we assume that f(*) has derivative bounded in $(0,\infty)$ and absolutely integrable in $[0,\infty)$.

For these reasons what i wrote is valid only under the conditions reported by Gluskin...

Kind regards

$\chi$ $\sigma$