$f$ is real valued function on $[0,\infty)$ such that $f''(x)&gt;0$ for all $x$ and $f(0)=0$  Then $h(x)={f(x)\over x}$ is decreasing or increasing or both over $[0,\infty)$  $h'(x)={xf'(x)-f(x)\over x^2}$  What I can conclude from here?
122170
Suppose $f(0) =0 $ and  $0&lt;f''(x)&lt;\infty (\forall$  $x&gt;0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases.    I have shown that $f'(x)-\frac{f(x)}{x}  = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here?
119987
Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$?  If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.
93108
If $f(x) \in C(-\infty+\infty)$, $\;g(x)=f(x) \int_0^x f(t)\,dt\,$ and $\;g(x)$ is monotone-decreasing in $(-\infty,+\infty),$  Prove:$f(x)\equiv 0$.  It is easy to get $g(0)=0$,and I'm thinking about proving $g(x)\equiv 0$,then $f(x)\equiv 0$ follows.  But I havn't any ideas to prove $g(x)\equiv 0$.
49670
Show $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$   Proof:  By definition:    $$f'(x) = \lim_{h \to \ 0} \frac{f(x + h) - f(x)}{h}$$  Using this idea it would imply:   $$1)\ \ f''(x) = \lim_{h \to \ 0} \frac{f'(x + h) - f'(x)}{h}$$  As such it is required that I find an expression for $f'(x+h)$. This is where I'm not sure if the step I took is legitimate.  An expression for $f'(x + h)$ is:  $$f'(x+h) =  \lim_{h \to \ 0} \frac{f(x + 2h) - f(x + h)}{h}$$  Combining this with the definition of $f'(x)$ and inserting it into 1) you arrive at: $$\lim_{h \to \ 0} \frac{f(x + 2h) - 2f(x+h) + f(x)}{h^{2}} = f''(x)$$   As required.    Concern: I feel a discomfort with this solution. Even though "mechanically" it worked out, if I am taking the limit as $h \rightarrow 0$ that would mean $x + 2h$ and $x + h$ both go to $x$. But I am attempting to use the idea that $x +2h$ goes to $x + h$. Perhaps it is a notation idea that I need to communicate better, but I feel it is larger than just that.
88696
Suppose the function $f:[0,\delta) \to \mathbb{R}$ is continuous, differentiable in $(0,\delta)$ and $f(0)=0$.  If the limit $\displaystyle \lim_{x \to 0+}\frac{f(x)}{f'(x)}= L$ exists, then is it always the case that $L = 0$.  This seems to be true both for functions with well-behaved derivatives such as $f(x) = x,$  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x}{1}=0,$$  as well as functions with "bad" derivatives such as   $$f(x) = \begin{cases}x \ln x &amp;\mbox{if }x&gt;0 ,\\0  &amp;\mbox{if } x=0,  \end{cases},$$  where  $$\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.$$  Is there a simple proof or counterexample?
89554
Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x &gt; 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help?
99523
The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$
42395
If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
23109
Let $f:\mathbb{R}\to\mathrm{S}$, where $\mathrm{S}\subseteq\mathbb{R}$, be a strictly increasing (decreasing) function. I wish to find out whether its inverse $f^{-1}$ is also strictly increasing (decreasing) or not. While a proof for this exists online, I will provide a counterexample that seems to contradict the conclusions in the proof.  The proof can be found in:   I will for simplicity only consider a strictly increasing function. From the provided link above we have that: \begin{align*} x &lt; y &amp; \quad\Leftrightarrow\quad f(x) &lt; f(y) \\ f^{-1}(x) &lt; f^{-1}(y) &amp; \quad\Leftrightarrow\quad f^{-1}(f(x)) &lt; f^{-1}(f(y)) \\ f^{-1}(x) &lt; f^{-1}(y) &amp; \quad\Leftrightarrow\quad x &lt; y \\ \end{align*} Thus, if $f$ is a strictly increasing function, then so is $f^{-1}$.  Now for the counterexample. With some loss of generality, let $f:\mathbb{R}\to\mathbb{R}_{++}$ be a strictly positive function, where $\mathbb{R}_{++}=\{ x\in\mathbb{R}: x&gt;0\}$. Furthermore, for an inverse it follows that $f^{-1}(x)f(x)=f(x)f^{-1}(x)=1$. Then from  \begin{align*} x &lt; y &amp; \quad\Leftrightarrow\quad f(x) &lt; f(y) \end{align*} we can rewrite the right-hand side as  \begin{align*} f^{-1}(y)=\frac{1}{f(y)}&lt;\frac{1}{f(x)}=f^{-1}(x) \end{align*} which implies that if $f$ is a strictly increasing function, then its inverse $f^{-1}$ is strictly decreasing.  Please provide comments if and why my counterexample is invalid. Thank you in advance.
11660
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance.
120426
I am really satisfied that $(x \to a) ≠(x=a)$ and if that is not right , Then all the process of $Limits$ is dividing by zero and that is a crime.  Since $(x \to a) + h = (x=a)$ , $h ≠ 0$,So Why does $f(x \to a) = f(x = a)$ ?  NOTE:I am talking about continuous function.
19294
Let $f: \Bbb R \to \Bbb R$ be a function such that $f'(x)$ exists and is continuous over $\Bbb R$. Moreover, let there be a $T &gt; 0$ such that $f(x + T) = f(x)$ for all $x \in \Bbb R$ and let $f(x) + f'(x)\ge 0$ for all $x \in \Bbb R$.      Show that $f(x) \ge 0$ for all $x \in \Bbb R$.   My attempt: $f(x) \ge 0 \iff f(x) \ge f'(x) - f'(x) \iff f(x) + f'(x) \ge f'(x)$.  Thus, it is enoguh to show that $0 \ge f'(x)$.  $\iff 0 \ge \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$  I do not know how to proceed from here. I know that $f'$ also has a periodicity of T but I do not know how to use that here.  Am I on the right track? How can I use the periodicity of $f$ to solve the problem?
59864
Let $l \in\mathbb{R} $ and $ f:(0,\infty) \to \Bbb R $ be a function such that $ \lim_{x\to \infty} xf(x)=l $ .  Prove that  $ \lim_{x\to \infty} f(x)=0 $.  Any help would be appreciated - I found this difficult to prove
93339
Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $,  $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.  $\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq  K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea.
891
Does $f(f(f(x)))=x \implies  f(x)=x$? Is it necessary for it to follow? How do we prove this? Do we have to substitute some special $x$? Or is it some other consideration of its properties?  Edit: does it follow in the special case when f is surjective? Or would we wind up with the same thing?
51766
If $f'(c)$ exists then $\lim_{x\to c} f(x)=f(c)$   My attempt is below. Did I do wrong? Since I don't really understand it. I thought $f'(x) = f'(c)$ so $f'(x)-f'(c)=0$. Am I on the right track? Many thanks!  \begin{align} f(x)-f(c)&amp;=\frac{f(x)-f(c)}{x-c}\cdot(x-c) \\ \lim_{x\to c}[f(x)-f(c)]&amp;=\lim_{x\to c}\left[\frac{f(x)-f(c)}{x-c}\cdot(x-c)\right] \\ \lim_{x\to c}[f(x)-f(c)]&amp;=0 \\ \lim_{x\to c} f(x)&amp;=f(c) \end{align}
74431
I've been stuck on this question for a long time now and was wondering if anyone could show me how it's done. So far I have done the following: Since $\lim_{x\to \infty}f'(x)=M$ then $\forall \epsilon ,\exists A &gt;0$  s.t. if $x&gt;A$ then $|f'(x)-M|&lt;\epsilon$. So we have that $M-\epsilon&lt;f'(x)&lt;M+\epsilon$. By MVT on $[x,x+1]$ for $x&gt;A$, $\exists c\in (x,x+1)$ s.t. ${f(x+1)-f(x)\over 1}=f'(c)$. Therefore, $M-\epsilon&lt;f(x+1)-f(x)&lt;M+\epsilon$. However, I do not know what to do after this.