Suppose $f(x)$ is bounded and differentiable on $[0,\infty)$. Consider the statement: If $\lim_{x\to \infty}f(x)=0 $, then $\lim_{x\to \infty}f'(x)=0 $. Prove it if it's right or show a counterexample.  I think it's not true and trying to find a counterexample by using functions containing $\cos(\frac{1}{x})$ or $e^{-x}$ or other things. The final goal is to find something make it's derivative' limit at infite doesn't exist. Since if it exists, it must equals zero or $f(x)$ will not be bounded.
49415
I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L &lt; \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?  It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:  $$ \begin{align} \lim_{x \to +\infty} f'(x) &amp;= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &amp;= \lim_{h \to 0} \frac1{h}(L-L) \\ &amp;= \lim_{h \to 0} \frac{0}{h} \\ &amp;= 0 \end{align} $$  In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is  already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case.
106246
Problem: Let $g:[0,1]\to\mathbb{R}$ be a continuos function such that $g(1)=0$. Show that $\sup\limits_{x\in[0,1]}|x^ng(x)|\to0$ as $n\to\infty$.  My approach: Let $f(x)=x^ng(x)$. Then, $f(x)$ is continuos on $[0,1]$. Using the extreme value theorem we can say, that $|f(x)|$ is bounded on $[0,1]$ and attains its bounds, i.e., $\exists x_0\in[0,1]$ such that $|f(x_0)|=\sup\limits_{x\in[0,1]}|f(x)|$.  Now, $|f(x_0)|=|x_0^ng(x_0)|$. If $x_0=0$, then we are done. So, let us assume $x_0\neq0$. By the Archimedean property of $\mathbb{R}$, $\exists$ a $y\in\mathbb{R}$ such that $y=x_0^n\leq1$, since $x_0\in[0,1]$.   Hence, $\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|x^ng(x)|=\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|f(x)|=\lim\limits_{n\to\infty}|f(x_0)|=\lim\limits_{n\to\infty}|yg(y^{\frac{1}{n}})|\leq\lim\limits_{n\to\infty}|g(y^{\frac{1}{n}})|=g(1)=0.$  I will be glad if anyone can tell me if I have done any mistake and if so please help me to correct it.  Thank you.
86812
The question I'm trying to answer is:  Let $I$ be an interval and let $f:I\rightarrow\mathbb{R}$ be differentiable on $I$. Show that if the derivative $f'$ is never 0 on $I$, then either $f'(x)&gt;0$ for all $x$ or $f'(x)&lt;0$ for all $x$.  I know I can apply Darboux's Theorem to solve the question.  My Attempt: Let $I=[a,b]$. If $k$ is a number s.t. $f'(a)&lt;k&lt;f'(b)$, then there exists a point $c$ in $(a,b)$ s.t. $f'(c)=k$. Because $f'\neq0$, $0&lt;f'(a)&lt;k&lt;f'(b)$.  I'm really stuck.. Could somebody help me out here? Thank you.
45796
Let $f:(0,\infty) \to R$ be differentiable. Suppose that $\lim_{x\to\infty}(f(x)+f'(x))=L$. Show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$. (Hint: Write $f(x) = e^xf(x)/e^x$ and use l’Hopital’s Rule.)    My working for $\lim_{x\to\infty}f'(x)=0$:  For $\lim_{x\to\infty}f'(x) = 0$, I let $f(x) = e^xf(x)/e^x$ and applied quotient rule which then cancels off $e^{2x}$ and I'm left with $\lim_{x\to\infty}(f(x)+f'(x)-f(x))$. Can I then equate this with $\lim_{x\to\infty} \left(f(x)+f'(x)\right) - \lim_{x\to\infty}f(x)$ which then gives $L-L=0$? Is this step correct?
23306
Let $f : \mathbb{R} \to \mathbb{R}$ be a function which is differentiable everywhere on $\mathbb{R}$. If we know that  $$\lim_{x\to-\infty}f(x) = L = \lim_{x\to\infty}f(x)$$ How would one go about showing that there must exist $c \in \mathbb{R}$, such that $f'(c) = 0$ for all cases at once?
30284
How to show that $\lim_{x \to +\infty}(f(x)+f'(x))=0 $ implies $\lim_{x \to +\infty} f(x)=0$?
107611
Why is $\lim_{x\to\infty} \frac{1}{x}$ equal to $0$?
45476
I'm trying to solve this question:     Let $f:[0,+\infty) \to \mathbb{R}$ be derivable and $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x\to \infty}\frac {f(x)}{x}=L$.   I'm trying to solve this question using l'Hôpital rule, but I couldn't use it, because I don't know if $\lim_{x\to +\infty} f(x)=+\infty$.  I need help.  Thanks a lot.
45139
Suppose for real numbers $a&lt;b$ one has a function with continuous derivative $f:[a,b] \to \mathbb{R}$ such that $f(a) = 0$ and there exists a real number $C$ with $$|f'(x)| \leq C|f(x)|$$ for all $x \in [a,b]$. Show that $f(x) = 0$ for all $x \in [a,b]$.  Well, since $f(a) = 0$, we have that $|f'(a)| \leq C|0|$, so $|f'(a)| = 0$. Since $f$ has a continuous derivative, we also know that $f$ is continuous. Since $f$ is continuous on a compact interval, $f$ obtains a maximum, say at $\xi \in [a,b]$. So, $|f'(x)| \leq C|f(\xi)|$. Since the derivative is bounded, we obtain that $f$ is Lipschitz, so $f$ is also uniformly continuous. Suppose to the contrary that $f(\xi) &gt; 0$? The above is pretty much everything I could figure out about $f$, so I'm not sure what to try next. This one is also from an old qual and possibly uses methods from beyond our course. I think maybe I should try to show that $|f'(x)| = 0$ for all $x$, but I don't know how.
87979
Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.   Please, see if I made some mistake in the proof below. I mention some theorems in the proof:   The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.  (1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.   (2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.   (3) We'll do this by the Precise Limit Definion: given $ \epsilon&gt;0, \exists\delta|0&lt;|x-x_0|&lt;\delta$, then $0&lt;|f(x)-f(x_0)|&lt;\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.
50361
Let $f:[0,\infty)\to [0,\infty)$ be a continuous function such that $\displaystyle \int_0^{\infty}f(t)\,dt&lt;\infty$. Which of the following statements are true ?   (A) The sequence $\{f(n)\}$ is bounded.  (B) $f(n)\to 0$ as $n \to \infty$.  (C) The series $\displaystyle \sum f(n)$ is convergent.  I am unable to prove directly but I am thinking about the function $f(x)=\frac{1}{1+x^2}$. For this function all options are correct. Is it correct ? I think not , as I have no proof in general.  Please help by giving a proof or disprove the statements.
13545
Problem:   If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.   This approach is definitely wrong:    \begin{align} b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &amp;=af'(0)-f'(0)\\ &amp;=(a-1)f'(0) \end{align}  I will show you a case why this approach is wrong:     \[f(x)= \begin{cases} 1,&amp;x\neq0\\ 0,&amp;x=0 \end{cases}\]   $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$   but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$   Does anyone know how to prove it? Thanks in advance!
70412
I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ does not exist. How precise is this statement knowing that this limit is $\infty$?. I thought saying the limit does not exist is not true where limits are $\infty$. But it is said when a function does not have a limit at all like  $$\lim_{x\to \infty}{\cos x}$$.
98512
I want to show that if $f$ is non-increasing and $f\in L_{1}([a,\infty),m)$ where $m$ is Lebesgue measure then $\lim_{t\to\infty} t f(t)=0$. So far I've been able to show that $f\geq 0$ and that $\lim_{t\to\infty} f(t)=0$. Since monotone functions are differentiable a.e. I thought about using integration by parts but couldn't get anywhere with that. Any hints or suggestions would be greatly appreciated.
54352
Having some trouble proving this:     If $f:\mathbb{R}\rightarrow \mathbb{R}$ is differentiable and $\lim_{x\to \infty } f^\prime(x)=0$ show: $\lim _{x\to \infty } (f(x+1)-f(x))=0$   Attempt: from $\int \lim _{x\rightarrow \infty } f^\prime(x) \mathrm{d}x=0$ we can say that $f^\prime(x)=const$ would we then have to show that  $\lim _{x\to \infty } (f(x+1))$ is also a constant? From there if we show that both are equal then the difference is zero. I am on the right track?  Thank you for the help.
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.
26952
Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by  to a different question)