Let $g:[1, \infty)\rightarrow \mathbb R$ a continuous non-negative function, such that $\int_{1}^{\infty} g(x)\ dx$ converges. Is it true that $\lim\limits_{x\to \infty}g(x)=0$ ?   I tried to find a counter-example but I can't figure a trivial one. I also tried to prove it by the definition of the convergence of $g(x)$ but couldn't show that limit is really $0$.
8851
Is there a continuous function such that $\int_0^{\infty} f(x)dx$ converges, yet $\lim_{x\rightarrow \infty}f(x) \ne 0$?  I know there are such functions, but I just can't think of any example.
117760
Consider the sequence of functions  $$ f_n(x) = \frac{x^2}{n} $$ for $x \in \mathbb{R}$  I have shown that $f_n(x)$ converges pointwise towards $0$ for $n \rightarrow \infty$ but I am not sure whether my attempt for showing that it does not converge uniformly towards $0$ for $n \rightarrow \infty$ is correct. Do you mind verifying?   By negation we have that $f_n(x)$ does not converge towards $0$ for $n \rightarrow \infty$ if   $$\exists \epsilon &gt; 0 \forall n \in \mathbb{N} \exists x \in \mathbb{R} \exists n \in \mathbb{N}: n \geq N \Rightarrow |\frac{x^2}{n}| \geq \epsilon$$  Let $\epsilon = 1$. Then for all $n \in \mathbb{N}$ we can find a $x \in \mathbb{R}$ such that for $n \geq N$ that $|x^2/n| \geq 1$. Can I then just pick $x = n^{1/2}$ so we have that $$ |x^2/n| = |(n^{1/2})^2/n| = |n/n| = 1 \geq 1 $$ which means that $f_n(x)$ does not converge uniformly towards $0$ for $n \rightarrow \infty$. Is this ok?
21229
Please help me out with this problem:  Let $f_n(x) : [0,1] \to \mathbb R$ be a sequence of continuous functions convergent at every $x \in [0,1]$ to a continuous function $f: [0,1]  \to \mathbb R$ . Does $f_n$ converge uniformly to $f$?  My first solution: I read this example in  posted earlier and I think it works: Take a sequence $f_n:[0,1] \to \mathbb R$ such that $f_n$ increases linearly from $0$ to $1$ on the interval $\left[ 0,\frac{1}{n} \right]$, decreases linearly from $1$ to $0$ on the intreval $\left[ \frac{1}{n}, \frac{2}{n} \right]$, and is $0$ on $\left[ \frac{2}{n},1 \right]$. Then $f_{n} \to 0$ nonuniformly.  Is this answer true? Also, if anyone can give some examples of such function with detailed proof, I will very appreciative?
59694
Let $g:[0,1] \rightarrow \mathbb R$ be a continuous function and assume that    $$ \int_{0}^{1} g(x) \phi'(x) dx = 0 $$   for all continuously differentiable functions $\phi: [0,1] \rightarrow \mathbb R$, where $\phi (0) = \phi(1) = 0$. Show that $g(x) = c$ for all $x \in [0,1]$ for some constant $c \in \mathbb R$.    My thoughts are applying integration by parts with $G(t) = \int_{0}^{t} g(x) dx $, since now we get  $$ 0 = \int_{0}^{1} G(t) \phi'(t) dt  + \int_{0}^{1} g(t) \phi (t) dt $$  which doesn't really go far. The second thought was applying MVT for definite integrals, which doesn't go far either as the conditions are hardly satisfied.   Any hints?
64509
Question:  a. Find an example for a non-negative and continuous function s.t. $\int _0^\infty f(x)dx$ is finite but the following limit doesn't exist: $\lim_{x\to \infty} f(x)$.  b. Is it possible that $\int _0^\infty f(x)dx$ is bounded but $f(x)$ is not bounded?  What we did with A: We suggested the function $|sin(x^2)|$ which has periods that get smaller and smaller until they no longer imply on the sum. But since it's a trig function it doesn't have a determinate limit. Wolfram didn't have the integral value for that func, and we were wondering if it really converges and if it is really a good example.  What we did with B: We thought about the function: $f(x)=  { x\in \Bbb N },{e^{-x} \notin \Bbb N }$ and we had a disagreement whether f(x) is integrable. I said no because similarly to Dirichlet function, one sum's limit will be 0 while the other one's will be infinite. My partner said that if I find a $\delta&lt;1$ then the definition of Riemann's integral does hold and so this integral is equal to that of $f(x)=e^{-x}$ She was trying to say this functions integral will be 0, but still it won't be bounded. I disagreed. Who's right?
94924
Show that if $f\in L^2 (\mathbb{R})$ then    $$\lim_{x \rightarrow \infty} g(x)=0,$$   where    $$g(x)=\int_x^{x+1}f(t)\,dt$$   Since $f \in  L^2 (\mathbb{R})$ then $f \in L^2[x,x+1]$ for every $x$. Then  $$|g(x)| \leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}} \int_x^{x+1} 1 dt$$ $$|g(x)|\leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}}$$
81277
I have this statement and I need to say if it is true or false:  Let $\{a_n\}$ be a real sequence.  $$\lim_{n\to +\infty} a_n = 0 \quad \implies \quad \sum_{n=1}^{\infty}(-1)^na_n \quad \text{converges}$$  I know, from the Leibniz criterion that:     If $a_n \to 0$, $a_n$ is decreasing and positive then $\sum_{n=1}^{\infty}(-1)^na_n$ converges   From this fact I believe that the statement is false but I couldn't come up with an infinitesimal sequence that isn't decreasing and for that reason is divergent. I tried some function with $sin(\frac{1}{n})$ without any luck. Any help would be very appreciated, thank you!
20693
Let $f:[0,1] \longrightarrow \mathbb R$ be a continuous function, and let $g:\mathbb R \longrightarrow \mathbb R$ be a continuous and periodic function with period $1$. Prove that   $\displaystyle\lim_{n\to \infty}\int_0^1f(x)g(nx)dx=\left(\int_0^1f(x)dx\right)\left(\int_0^1g(x)dx\right)$.  any Ideas?
103504
I was wondering if there is a function $f: \mathbb{R} \to \mathbb{R}$ such that its limit at every point is infinite.   I guess not, because what would its graph look like, but then again, I don't know how to prove it.
75574
I have some problems understanding  of this problem:  Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that  $$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$  It says there that if we split this integral into $n$ parts to get  $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$  we will have   $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$  because for large $n$, $f$ is roughly constant over the range.  Could you tell me why $f$ is roughly constant over the range for $n$ large enough?  I would really appreciate all your insight.
29696
I'm trying to solve this previous qual problem from my univeristy: Let $L_n$ be the continuous linear functions on $L^\infty(\mathbb{R})$ given by $$L_n(\phi)=\frac{1}{n!}\int_0^\infty x^ne^{-x}\phi(x)\,dx.$$  Prove that $L_n$ has no subsequence that converges in the weak* topology of $L^\infty(\mathbb{R})^\ast$. So far, I managed to see that $\frac{1}{n!}x^ne^{-x}$ converges uniformly on compact sets to 0, hence the weak* limit of any subsequence of $L_n$ must take compactly supported $L^\infty$ functions to 0. I've also tried to construct a test function directly for any subsequence but so far I have yet to come up with such a construction. Can anyone pleas help me? Some hints would be very much appreciated! Thanks!
26410
Let $f(x)$ be continuous except at $x = 0$ and let $a &gt; 0$. Assume that the improper integral $$\int_{0}^{a}f(x)\,dx = \lim_{\epsilon \to 0+}\int_{\epsilon}^{a}f(x)\,dx$$ exists and let $$g(x) = \int_{x}^{a}\frac{f(t)}{t}\,dt$$ Show that $$\int_{0}^{a}g(x)\,dx = \int_{0}^{a}f(x)\,dx$$  I tried integration by parts noting that $g'(x) = -f(x)/x$ and obtained for $0 &lt; \epsilon &lt; a$ the following $$\int_{\epsilon}^{a}g(x)\,dx = [xg(x)]_{x = \epsilon}^{x = a} - \int_{\epsilon}^{a}xg'(x)\,dx$$ or $$\int_{\epsilon}^{a}g(x)\,dx = -\epsilon g(\epsilon) + \int_{\epsilon}^{a}f(x)\,dx$$  The problem is solved if we can somehow show that $\lim_{\epsilon \to 0+}\epsilon g(\epsilon) = 0$. Looking at the definition $g(x)$ we see that we have no information of the behavior of $f(t)$ at $t = 0$ and the $t$ in denominator complicates the analysis of $g(\epsilon)$. Please suggest some hints which can lead to the solution.  Note: This problem is taken from G. H. Hardy's "A Course of Pure Mathematics" 10th ed. Page 397.
109939
Let $f$ be a continuous function. I wish to show $$\lim_{n \rightarrow \infty} \int_0^1 n x^n f(x) \; dx = f(1)$$ I can try to split up the integral over intervals $[0, 1-\delta]$, $[1-\delta, 1]$. The integral over $[0, 1-\delta]$ vanishes as $n \rightarrow \infty$, but I am having difficulty estimating the integral over $[1-\delta, 1]$. Thanks!
23237
Let $(a_n)_n$ be a sequence of non-negative real numbers.      Prove that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges.   I don't necessarily need a full solution, just a clue on how to proceed would be appreciated.  Could any convergence test like the ratio or the root test help?
6398
How does one show that $\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x$ converges? I have a book that uses this to define $e$ but they don't show that it converges. I suppose we could give it an upper and lower bound but I don't know how to do this. I've seen this question before on here but I haven't seen an answer I like yet.
103453
If $\lim_{x \to a}f(x)=1$ and $\lim_{x\to a}g(x)=\infty$, then  $$\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x\to a}g(x)(f(x)-1)}.$$ How can one prove this theorem?
81593
Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that  $$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$  exists and compute the limit.  Attempt:   Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval  and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&amp;=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &amp;=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &amp;=\int_0^1 f(0)dx=1. \end{align*}  Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.  My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified.