If $f:R\to R$ is bounded on bounded intervals, and $\lim f(x+1) - f(x) = L$ when $x\to +\infty$ then $\lim f(x)/x = L$ when $x\to +\infty$   I found answers here at the site when $f$ is differentiable, but I just don't know what to use in this case, where we don't have differentiability (at least not explicitly stated). Any help would be appreciated.
2876
Let $f:[0,+\infty)\to\Bbb R$ be a function bounded on each finite interval. I want to show that if $\lim\limits_{x\to+\infty}[f(x+1)-f(x)]= L,$ then also $\lim\limits_{x\to+\infty}\dfrac{f(x)}x = L$
65691
I'm looking for a good proof for the following problem. I think it's pretty intuitive why this is true, but I'm having trouble expressing it rigorously.     Suppose that $f$ is twice differentiable and that $f''$ is bounded on   $(0, \infty)$. Furthermore, suppose that $\lim_{x\to \infty} f(x)=0$.   Prove that $\lim_{x\to\infty} f'(x)=0$.   I've already tried writing out the Taylor expansion for $f$ and tweaking that, but I haven't had much success. Any insight would be greatly appreciated.
61669
Show that $|x|&lt;1$ implies that $\lim_{n \to \infty}x^n$ = 0.   Can anyone help me define my $N$ (as used in the formal definition of a limit) in terms of $\varepsilon$? I define $x$ as $\dfrac{1}{1+p}$ where $p&gt;0$ but I am not sure how also to define $N$ since both $P$ and $\varepsilon$ are variable.
92698
I conjecture that in some specific conditions a differentiating function gives the following equality: $$\lim_{n\to\infty} f(n+1) - f(n) = \lim_{x\to\infty} (f(x))' $$  However, I'm not sure yet what exactly those conditions are in order to precisely know where I may apply this rule or not. If you wanna take a look over my posted problem  you'll immediately notice that this rule applies for that case. I really appreciate if you help me clarify this.
23404
If $$\lim_{x\to\infty}(f(x)+f'(x))=l$$ then prove that $$\lim_{x\to\infty}f(x)=l \text{ and } \lim_{x\to\infty}f'(x)=0 $$    I assume four cases $$\begin{array}{c|c|c|c|}   &amp; f(x) &amp; f'(x) \\ \hline \text{1} &amp; \infty &amp; -\infty \\ \hline \text{2} &amp; -\infty &amp; \infty \\ \hline \text{3} &amp; l &amp; 0 \\ \hline \text{4} &amp; 0 &amp; l \\ \hline \end{array}$$ and elimination(1,2,4) of not possible cases can give the answer(3).    My work is not a correct/perfect or flaw proof.What would be a correct one or is this correct.
79487
Let $f$ be differentiable on $\Bbb{R}$ and let $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$. Prove there exists $a\in \Bbb{R}$ such that $f'(a)=0$.  Attempt: If $f$ is constant we are done. Otherwise, $f$ is not constant. Suppose there is no such $a$. Then, in particular, $f$ has no extremum $\implies$ $f$ has no point where is goes from increasing to decreasing or from decreasing to increasing. $f$ is continuous since it is differentiable. Hence, since $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$, $f$ must be constant. A contradiction. Therefore either $f$ is constant and $f'(x)=0$ everywhere, or $f$ has an extremum at $x=a$ and $f'(a)=0$.   I feel like my attempt is not formal enough, any correction?  I would appreciate your help.
15146
I want to show that for $f: \mathbb{R} \rightarrow\mathbb{R}$ Lebesgue integrable  $\displaystyle{\lim_{n \to +\infty} f(x+n) =0}$ almost everywhere.  It is clear to me that this statement is true and I have tried to prove it by contradiction, by assuming that either $\displaystyle{\lim_{n \to +\infty} f(x+n) =\infty}$ or$\displaystyle{\lim_{n \to +\infty} f(x+n) =K}$ with $K\in \mathbb{R}$ .  In the first case  we can follow that $|f|=\infty$ so that $\int_\mathbb{R}f=\infty$ and we have a contradiction.  In the other case we know that there must be an $a\in \mathbb{R}$ so that $\forall x&gt;a$ we have $f_+(x)&gt;K/2$ and by that:  $\int_{{\mathbb{R}}+}|f|&gt;\int_{\mathbb{R}_+}f_+=\int_{[0,a]}f_++ \int_{(a,\infty)}f_+&gt;\int_{[0,a]}f_++\int_{(a,\infty)}K/2=\infty$  Would this be enough to show the statement? Can I even make the assumption that the limit of $f(x+n)$ "exists"? Any hints would be great.
16048
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
103954
If $f$ Continuous bounded on $\mathbb{R}$，and $$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$ prove that $f$ uniformly continuous on $\mathbb{R}$. I try to find a $\delta&gt;0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|&lt;\delta$have$$\left|f(x_1)-f(x_2)\right|&lt;\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!
78312
Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
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My teacher proposed this question as a challenge proof to do on our own and I can't seem to get it. Was Wondering if anyone could give me a hint or help me on the process to completing it.  Let $a \in \mathbb{R}$. Let $f$ be a function defined, at least, on an interval  centred at $a$, except possibly at $a$. Let $L \in \mathbb{R}$.     If $\lim\limits_{ x→a-} f(x) = \lim\limits_{ x→a+} f(x) = L$   then $\lim\limits_{ x→a} f(x) $ = $L$.    We are supposed to prove this using the "delta-epsilon" proof style. Thanks in advance!!
123499
Let $f:[0, +\infty) \rightarrow \mathbb{R} $ be a bounded function in each bounded interval. If $$\lim_{x \to +\infty} [f(x+1) - f(x)] = L$$ then $$\lim_{x \to +\infty} \frac{f(x)}{x} = L$$   I tried using the definition on the first limit; then, I attempted to use that inequality to apply the triangular inequality to arrive at the definition that gives me the second limit. It is supposed to be simple, but I'm not feeling safe as in how I should write it.
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.
77964
Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let $f(x)$ be bounded on any interval $(1,b), b&gt;1$. Then $\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))$. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure.
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Given $f:R \to R$, $f$ is differentiable on $R$ and $\lim_{x \to \infty}(f(x)-f(-x))=0$. I need to show that there is $x_0 \in R$ such that $f'(x_0)=0$  I am trying to prove it by contradiction .... so i assume there is no $x_0 \in R$ such that  $f'(x_0)=0$ this means this function is strictly monotonic because the derivative "respects" the mean value theorem so it cannot be negative and positive without passing a zero value ....  (from what i understand the derivative function of a continuous/differentiable functions isn't necessarily continuous but it still respect the mean value theorem i think it's called darboux's theorem )  now from $\lim_{x \to \infty}(f(x)-f(-x))=0$ ,we get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$  now the last step is the contradiction its very intuitive and even maybe obvious that we can't get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$ for strictly monotonic function but i cant think of a formal way to prove it .
11394
Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.   I can't think of a sequence to compare this to that'll work.  L'Hopital's rule doesn't apply.  I know there's some simple way of doing this, but it just isn't coming to me. :(
40274
I am finding difficulties in finding a counterexample to the following statement. If $f\colon (0,\infty) \to(0,\infty) $ is uniformly continuous, then $$\lim_{x\to \infty} \frac{f(x+\frac{1}{x})}{f(x)} =1.$$