Let $f:(0,\infty)\rightarrow\mathbb{R}$ be differentiable and let $b$ be a real number. Prove that if $f'(t)\rightarrow b$ as $t\rightarrow\infty$, then $\frac{f(t)}{t}\rightarrow b$ as $t\rightarrow\infty$.  Median Value Theorem: If $f$ is continuous and differentiable, then $\exists c$ such that $a&lt;c&lt;b$ and $f'(c)=\frac{f(b)-f(b)}{b-a}$. So when $c\rightarrow\infty$, $a,b\rightarrow\infty$. How do I get $\frac{f(t)}{t}$?
2385
If $f$ is differentiable on $(a,\infty)$, Show that if $f'(x)\rightarrow L$ as $ x \rightarrow \infty$, $-\infty \leq L \leq \infty $  then  $ f(x)/x \rightarrow L $ as $x \rightarrow \infty$.   Deduce further that if $f(x) \rightarrow M$ as $ x\rightarrow \infty$ , and $ f'(x)\rightarrow L $ as $ x \rightarrow\infty $ then $ L = 0$   For the first part I tried converting $f(x)$ to $F(\frac{1}{x})$ and $x \rightarrow 0$ and then using L'Hospital's rule but I am getting stuck. How do I proceed with this problem ?
17222
I am wondering if the following converse (or modification) of the mean value theorem holds. Suppose $f(\cdot)$ is continuously differentiable on $[a,b]$. Then for all $c \in (a,b)$ there exists $x$ and $y$ such that $$ f'(c)=\frac{f(y)-f(x)}{y-x} $$
47916
The problem:  Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $|f(t) - f(x)| \leq |t - x|^2$ for all $t, x$. Prove $f$ is constant.  I believe I have some intuition about why this is the case; i.e. if $t$ and $x$ are very close, ($|t - x| = \epsilon$), then $\sqrt{\epsilon}$ will converge to 0, so $|f(t) - f(x)|$ will also converge to zero if you keep taking $t$ and $x$ closer and closer to each other. However, how do I formalize this argument? Thanks.
29377
Let $f$ be differentiable and let $\lim_{x \to \infty}f'(x)=+\infty$ prove that: 1) $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and 2) $\lim_{x \to \infty}f(x)=+\infty$.  1) I'll prove by contradiction. Let $\lim_{x \to \infty}(f(x)-f(x-1))=a$, where $a$ is from and $\Bbb R$. $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=a$, where c is from $(x-1,x)$ (mean value theorem). As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=a$. Which is a contradictition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .   2) I will also prove this by contradiction. We have that $\lim_{x \to \infty}f(x)=a$ This is equivalent to $\lim_{x \to \infty}f(x)-f(x-1)=0$. I use mean value theorem and get $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=0$, where c is from $(x-1,x)$. As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=0$. Which is a contractition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .
101432
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.
98623
Let $f:(a,b)\rightarrow \mathbb{R}$.   $f$ satisfied the following property:  If $\forall x_{1},x_{0},x_{2}\in(a,b)$ and $x_{1}&lt;x_{0}&lt;x_{2};$then$\frac{f(x_{0})-f(x_{1})}{x_{0}-x_{1}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}.$   My question : whether the function $f\in C((a,b))?$    We can get  $\displaystyle\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}. $Let $\displaystyle g(x)=\frac{f(x)-f(x_{0})}{x-x_{0}},$ if we can prove $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0});$ then we have $f^{'}_{-}(x_{0}) $ exists .In a similar way ,$f^{'}_{+}(x_{0}) $ exists.  $f^{'}_{-}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}-}f(x)=f(x_{0});$ $f^{'}_{+}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}+}f(x)=f(x_{0}).$  Obviously, $f$ is coutinuous at $x=x_{0}.$Further, $f\in C((a,b))!$     But  I failed to prove  $g(x)$ is bound and  decreasing (not strictly ) in $(x_{0}-\delta ,x_{0}).$Sometimes  I doubt the conculsion that  $f\in C((a,b)).$  Either make a counterexample to deny it ,or prove the conculsion is correct?
30364
I know and understand the mean value theorem. But at the moment I don't have the intuition to understand the generalized mean value theorem     If $f$ and $g$ are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists a point $c\in(a,b)$ where$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$If $g'$ is never zero on $(a,b)$, then the conclusion can be stated as$$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.$$   What is the intuition? And how can you prove this (generally) ?
54805
Let $F:\mathbb{R^2}\rightarrow \mathbb{R}$ be a function, such as : $$F(x,y)=\begin{cases}        0 &amp; (x,y)=(0,0) \\       \frac{xy}{x^2+y^2} &amp; (x,y)\neq (0,0)    \end{cases}$$      Is $F$ differentiable at $(0,0)$?   My attempt :  Consider the limit, $\lim_{(x,y)\rightarrow (0,0)} F(x,y)$ along the line $y=x$. The limit equals to $\frac{1}{2}$. However, $F(0,0)=0$. Hence $F$ is not continuous at $(0,0)$ which implies that $F$ isn't differentiable as well.  Is this approach okay ?
84595
If we let $f_n$ be a sequence of functions on $[a,b]$ such that,  -there exists a point $x_0\in[a,b]$ where $lim_{n-&gt;\infty}f_n(x_0)$ exists -each $f_n$ is differentiable -the sequence $(f'_n)$ converges uniformly on $[a,b]$  I want to prove that $f_n$ converges uniformly on $[a,b]$ to some function $f$  My logic:  if I use the triangle inequality to write,  $$|f_n(x)-f_m(x)|&lt;=|(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|+|f_n(x_0)-f_m(x_0)|$$  Then I can try and apply the mean value theorem to the first term on the left hand side to prove that $f_n$ converges uniformly on $[a,b]$ to some function $f$.  The mean value theorem states,  Suppose f(x) is a function that satisfies both of the following conditions,  -$f(x)$ is continuous on the closed interval $[a,b]$ -$f(x)$ is differentiable on the open interbal $(a,b)$  Then there is a number $c$ such that $a&lt;c&lt;b$ and,  $f'(c)=\frac{f(b)-f(a)}{b-a}$
113206
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be smooth and such that for every $x$    $$\int_{-\infty}^{\infty}\frac{|f(x)-f(y)|}{|x-y|^2}dy &lt; \infty$$   Prove that $f$ is constant.
98940
I know that this can not be true due to counter-examples but I don't know where the error in my reasoning is.  Assumption: If $f(x)$ is differentiable in $\mathbb{R}$ then the derivative $f'(x)$ is continuous in  $\mathbb{R}$.  Faulty Proof: For every $c \in \mathbb{R}$, using the mean value theorem for $f(x),$ on the interval $x \in [c, c + h] $ where $h$ is positive.  $$ \frac{f(c + h) - f(c)}{h} = f'(\xi(h)) $$  Where $\xi(h) \in (c,c+h)$. Because this equation holds for every $h&gt;0$. It must hold in the limit as $h \rightarrow 0^+$.   $$ \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h))  $$  But the left side of the equation is the right one sided derivative.  $$ f'_{+}(c) = \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h))  $$  The same can be done for $h$ being negative, but because of differentiability at every point the left and right derivatives must be equal.    $$ f'(c) = f'_{+}(c) = f'_{-}(c) = \lim_{h\to 0^-}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^-}f'(\xi(h))  $$  As $h \rightarrow 0^+$, $\xi(h) \rightarrow c$. So because the limit $\lim_{h\to 0^+}f'(\xi(h))$ exists and $\xi(h) \neq c$, it is equal to $\lim_{x\to c^+}f'(x)$  It follows that $\lim_{x\to c^+}f'(x) = \lim_{x\to c^-}f'(x) = f'(c)$ so the function $f'(x)$ is continuous.
84290
Let $$f(x)=\begin{cases} \sin \tfrac 1 x &amp;\text{if $x\ne 0$}\\ 0 &amp; \text{if $x = 0$.}\end{cases}$$ I have to show that $f$ has the intermediate value property. That is, for any $a &lt; b$, if $y$ is any real number such that $f(a) &lt; y&lt; f(b)$ or $f(a)&gt;y&gt; f(b)$, then there exists a $c \in (a,b)$ such that $f(c)=y$.  I feel like I kind of know how to go about completing this. I just am curious as to if I have to create a bound such as letting $a = -1$ and $b = 1$, or keep $a$ and $b$ both arbitrary.
64075
$f:\Bbb R \to\Bbb R $ is differentiable, $f(0)=0$ and $|f'(x)|\le|f(x)|$ for all $x$ then prove $f$ is identically zero.  I tried to use mean value theorem and end up in $|f(x)|\le |x||f(c)|$ for some constant $c$ which depends on $x$. Can you help me little bit...
37886
Assume we have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that  $f$ is differentiable for all non-zero $x$, $f$ is continuous at $0$ and  $$\lim_{x \uparrow 0} f'(x) = \lim_{x \downarrow 0} f'(x) &lt; \infty. $$  Is it then true that $f$ is also differentiable at $0$?  If we drop the hypothesis that $f$ is continuous, this is not true since the signums function is a simple counterexample. But what if $f$ is continuous? I don't think this is true but I can't find a counterexample...
38575
I have a homework question, looks simple but I can't figure out a way to solve it. Any clue or help will be helpful.  Let $f: [0,\infty)\rightarrow\mathbb R$ be a differentiable function such that $\lim_{x\rightarrow\infty}f'(x)=0$.  Prove or disprove: $\lim_{x\rightarrow\infty}f(x)$ exists (infinite limit is also considered as a limit).  Thanks a lot!
89090
Can anyone help me with this problem?     Let $f:[0,\infty)\longrightarrow \mathbb R$ be a continuous function such that for each $x&gt;0$, we have $\lim_{n\to \infty}f(nx)=0$. Then prove that $\lim_{x\to \infty}f(x)=0$.   Our teacher told first to prove Baire's theorem, and then show that this is a consequence of that theorem. I proved Baire's theorem, and I spend a few hours thinking on how Baire's theorem is related to this problem, but I couldn't find anything. I'd really appreciate your help.
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.