Let $f:(a,\infty)\rightarrow \mathbb{R}$ be a function such that $f$ is bounded in any finite interval $(a,b]$. Prove that $\lim_{x \to \infty} f(x)/x=\lim_{x \to \infty}[f(x+1)-f(x)]$, provided that the right limit exists. Thanks a lot.
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.
84418
Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.  Suppose that $f:\mathbb N\rightarrow\mathbb R$. If $$\lim_{n\rightarrow\infty}f(n+1)-f(n) = L$$ prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.  This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.  Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.
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.
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!
48176
I'm having trouble with this homework question:  "Let there be a sequence $\left(b_{n}\right)_{n=1}^{\infty}  $ such that $\lim_{n\rightarrow\infty}\left(b_{n}\right)=1$. Let there also be some bounded sequence  $\left(a_{n}\right)_{n=1}^{\infty}$  Prove that the following statement is correct: $\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right)  $"  Alright so I sat a good two hours on this question with no major breakthroughs as of yet. I have some semi-proofs all of which use some arbitrary assumptions which I can't make.    Any hints?  Thank you!
6235
I don't know how to approach this exercise: Let $f$ be a real valued convex function that satisfies:  $$\lim_{x\to-\infty}f(x)=-\infty$$  prove that it must be $$\lim_{x\to+\infty}f(x)=+\infty$$  I know the definition of convexity and limits however I don't know how to prove this formally.
118491
Let $g:A\rightarrow \mathbb{R}$ and let $f$ be a bounded function on $A$. Show that if $\displaystyle\lim_{x \rightarrow c}g(x)=0$, then $\displaystyle\lim_{x \rightarrow c}g(x)f(x)=0$ as well.   Since $\displaystyle\lim_{x \rightarrow c}g(x)=0$, we know that $\forall \epsilon&gt;0 $ $\exists \delta &gt;0 $ so that whenever $0&lt;|x-c|&lt;\delta$, it follows that $|g(x)-L| &lt; \epsilon$. And $f(x)$ is bounded, then there is a number $M&gt;0$ satisfying $|f(x)|\leq M$ $\forall x \in A$  $f(x)$ is bounded it means all sequences in $f $ are less than $M$, so it doesn't seem like a stretch to say that   $|g(x)f(x) - L| &lt; \epsilon$, where, as the problem stated, $L=0$  Then $|g(x)| &lt; \frac{\epsilon} {M}$, since $M$ is an upperbound for $f$  And I'm not quite sure what to do next, any help would be greatly appreciated.
75107
I have this question :  Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so $f$ bounded.  I want to know if my proof is valid :  If continuous in $x_0$ then :  $$\lim_{x \to x_0} f(x)= f(x_0)$$  Therefore :  All $\epsilon&gt;0$ exists $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;\epsilon$.  Therefore for $\epsilon=1$ there is $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;1$.  $|x-x_0|&lt;\delta \rightarrow x_0-\delta&lt;x&lt;x_0+\delta$.  Lets choose $x_1,x_2$ that implies $x_0-\delta&lt;x_1&lt;x_2&lt;x_0+\delta$  Therefore in $[x_1,x_2]$ from Weierstrass from theorem bounded there.
106046
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a bounded function and suppose that $g(x)=\sup_{t&gt;x}f(t)$. Show that $\lim_{x\to a^{+}}g(x) = g(a)$ for all real $a$.   I have a hard time with these kind of proofs because in high school, I would simply replace $x$ by $a$ in $g(x)$ to get that $g(x) \rightarrow g(a)$. Obviously, Real Analysis requires a more formal proof.  I can always apply the definition of a limit of a function :  Given $\varepsilon&gt;0$, there exists $\delta&gt;0$ such that if $x\in A$ and $0&lt;|x-c|&lt;\delta$, then $|f(x)-L| &lt; \varepsilon$.  In particular, we have that if  $$ |x-c| = |x-a^{+}|&lt;\delta$$  then  $$|g(x)-g(a)| &lt; \varepsilon$$  How can I find a relation between $f$ and $g$, the sup of $f(t)$ and between $x$ and $a^{+}$?
14517
I have lecture notes with the claim $(C_b(X), \|\cdot\|_\infty)$, the space of bounded continuous functions with the sup norm is complete.  The lecturer then proved two things, (i) that $f(x) = \lim f_n (x)$ is bounded and (ii) that $\lim f_n \in \mathbb{R}$.  I don't understand why it's not enough that $f$ is bounded. I think the limit of a sequence of continuous functions is continuous and then if $f$ is bounded, it's in $C_b(X)$. So what is this $\lim f_n \in \mathbb{R}$ about? Many thanks for your help.
45963
1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
68123
I was trying to work out a problem I found online. Here is the problem statement:     Let $f(x)$ be continuously differentiable on $(0, \infty)$ and suppose $\lim\limits_{x \to \infty} f&#39;(x) = 0$. Prove that $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$.    (source: )  The first idea that came to my mind was to show that for all $\epsilon &gt; 0$, we have $|f(x)| &lt; \epsilon|x|$ for sufficiently large $x$. (And I believe I could do this using the fact that $f&#39;(x) \to 0$ as $x \to \infty$.)  However, I was wondering if there was a different (and nicer or cleverer) way. Here's an idea I had in mind:  If $f$ is bounded, then $\frac{f(x)}{x}$ clearly goes to zero. If $\lim\limits_{x \to \infty} f(x)$ is either $+\infty$ or $-\infty$, then we can apply l'Hôpital's rule (to get $\lim\limits_{x \to \infty} \frac{f(x)}{x} = \lim\limits_{x \to \infty} \frac{f'(x)}{1} = 0$).   However, I'm not sure what I could do in the remaining case (when $f$ is unbounded but oscillates like crazy). Is there a way to finish the proof from here?   Also, are there other ways of proving the given statement?
18467
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=\max\{|f(x)| \; :\; x \in [a,b]\}$. Is it true that: $$ M= \lim_{n\to\infty}\left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n} ? $$  Thanks!
22743
I am wondering whether there exists a function such that:  $$\lim_{x \rightarrow a}f(x)=\infty$$  at some point $a$ on the real axis but yet,  $$\int_{-\infty}^{+\infty}\left|f(x)\right|\ dx&lt;\infty$$  Does the fact that a function is unbounded imply that it has no finite integral?
21265
Suppose $f : [ 0, 1 ] → \mathbb{R}$ and $\lim\limits_{x \rightarrow c} f ( x )$ exists for all $c \in [ a , b ]$. Show that $f$ is Riemann integrable on $[ a , b ]$.  I can show this $f$ is bounded on $[a,b]$ since the limit exists at every point. But I'm not sure how to proceed. Any hints or help would be great.
14373
$$\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)$$  when f is continuous on $[0,1]$  I know it can be proved using bounded convergence theorem but,  I wanna know proof using only basic properties of riemann integral and fundamental theorem of calculus and MVT for integrals ...   Thank you.
66776
Let $(x_n)$ be a bounded sequence of real numbers such that $x_n \ge0$ for each $n \in \mathbb N$. Show that $\liminf(\sqrt{x_n})=\sqrt{\liminf x_n}$.   I assume we need to consider the subsequence of ($\sqrt{x_n}$), but I'm really confused with the  notion on "boundedness". Can I assume the question implies an upper bound on $(x_n)$?  Even then wouldn't the equation be implied? Given how roots generally affect limits.