real analysis – Given $f:mathbb{R} to mathbb{R}$ continuous with $f(f(x))=e^x$, show that $lim_{xto infty } frac{f(x)}{x^n}=infty$ for all $ninmathbb{N}$

Thanks for contributing an answer to Mathematics Stack Exchange!

  • Please be sure to answer the question. Provide details and share your research!

But avoid

  • Asking for help, clarification, or responding to other answers.
  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

calculus – What is $lim_{xto infty} sin^{-1}(x) / x$

Problem number 31 in section 6.8 in my version of Stewart’s Calculus book has the following limit
$$
lim_{xto infty} frac{sin^{-1}(x)}{x}
$$

My professor insists that this is equal to $1$. She says it is a simple application of L’Hopital’s Rule. So
$$
lim_{xto infty} frac{sin^{-1}(x)}{x} = lim_{xto infty} frac{1/sqrt{1-x^2}}{1}
$$

But this is $0$ I think.

Also, I isn’t the domain of the $sin^{-1}(x)$ just $(-1,1)$? How does this limit even make sense?

polynomials – How to evaluate $sqrt{5 +sqrt{5-sqrt{5 +sqrt{5 – sqrt{5 + …infty}}}}}$

The expression is as follows :

$x = sqrt{5 +sqrt{5-sqrt{5 +sqrt{5 – sqrt{5 + …infty}}}}}$

and we need to evaluate x.

My approach:

$x = sqrt{5 +sqrt{5-sqrt{5 +sqrt{5 – sqrt{5 + …infty}}}}}$

$x^2 = {5 +sqrt{5-sqrt{5 +sqrt{5 – sqrt{5 + …infty}}}}}$

$x^2 = {5 +sqrt{5-x}}$

$x^2 – 5 = sqrt{5-x}$

${(x^2 – 5)}^2 = {(sqrt{5-x})}^2$

$x^4 + 25 -10x^2 – 5 + x = 0$

$x^4 -10x^2 +x + 20 = 0$

Don’t know how to solve further. Please help. Different methods to approach this question would be appreciated. Thanks!

About $lim_{nto +infty} nprod_{k=1}^{n-1}cos^2(k)$

In my research work, I need to show that the sequence $(nu_n)$ tends to 0 where $ (u_n)$ is defined by $$u_{n+1}=u_{n} cos^{2}(n),quad u_{0}=1$$
$(u_n)$ is a positive and decreasing sequence. My adempt begin{align*}
u_{n+1}&= prod_{k=1}^ncos^2(k) =(prod_{k=1}^n frac{e^{k i } + e^{- k i }}{2} )^2 \
&= 4^{-n}e^{-2(1+2+cdots+n) cdot i } prod_{k=1}^n left( 1 + e^{2 k i } right)^2 \
&=4^{-n}e^{-n(n+1)cdot i } prod_{k=1}^n left( 1 + e^{2 k i } right)^2
end{align*}
Afterwards, I don’t see how to continue

functional analysis – Equicontinuity of $f_n(t) = left(e^{frac{t}{n}}, frac{cos(nt)}{n} , frac{t}{n}right)$ with $t in (-infty, 0)$

I’m trying to proof the equicontinuity of this sequence of functions. So, by definition, I want to proof that given any $t_0in(-infty,0)$

$$ forall varepsilon > 0, exists delta > 0 : forall ninmathbb{N}, forall tin(-infty,0), |t-t_0|<deltaRightarrow || f_n(t)-f_n(t_0)||<varepsilon$$

You can rewrite the last inequality as
$$left( e^{frac{t}{n}}-e^{frac{t_0}{n}}right)^2+left(frac{cos(nt)-cos(nt_0)}{n}right)^2+left(frac{t-t_0}{n}right)^2 <varepsilon^2$$

I’ve managed to bound the first and the third addend independently of n (uniformly), but I’m having some trouble trying to proof that
$$ exists delta_2>0 : forall ninmathbb{N}, |t-t_0|<delta_2 Rightarrow left(frac{cos(nt)-cos(nt_0)}{n}right)^2<frac{varepsilon^2}{3} $$

How to prove that $lim_{n to infty} x_{n} = x Longleftrightarrow x_{n} to x$ and $midmid x_{n} midmid to midmid x midmid$

How can I prove that $lim_{n to infty} x_{n} = x Longleftrightarrow x_{n} to x$ and $midmid x_{n} midmid to midmid x midmid$ when $n to infty$?

Show $limsup_{xto infty} sinleft((x+t)^2right)-sinleft(x^2right)=2$ for $ t ne 0$

My messing around on desmos strongly suggests to me that this is true, yet I have been been unable to prove it. I have tried to explicitly find a sequence $(x_n)$ such that $lim x_n =infty$ and $ lim sinleft((x_n+t)^2right)-sinleft(x_n^2right) = 2$ without success. Any help would be much appreciated.

$lim_{n rightarrow infty} frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1}$. Am I correct?

I have to find this limit:

begin{align} lim_{n rightarrow infty}
frac{(-1)^{n}sqrt{n}sin(n^{n})}{n+1} end{align}

My attempt:

Since we know that $-1leq sin (x) leq 1$ for all $x in mathbb{R}$ we have:

begin{align}
lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdot(-1)}{n+1}underbrace{leq}_{text{Is this fine?}}&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{(1)^{n}cdotsqrt{n}cdot(1)}{n+1}\ \ lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{n}}{n+1}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{n}}{n+1}
end{align}

My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have

begin{align}
lim_{n rightarrow infty} frac{(-1)^{n+1}cdotsqrt{frac{1}{n}}}{1+frac{1}{n}}leq&lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq lim_{n rightarrow infty} frac{sqrt{frac{1}{n}}}{1+frac{1}{n}}\ \ Rightarrow 0 leq &lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1} leq 0 \ \ therefore& lim_{n rightarrow infty} frac{(-1)^{n}cdotsqrt{n}cdotsin{(n^{n})}}{n+1}=0
end{align}

Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!

Problem proving $lim_{n rightarrow infty} int_S(f(x))^{frac{1}{n}}=mu(S)$

Statement:
Given a measurable set $S subseteq mathbb{R}$ and a measurable function $f:S longrightarrow mathbb{R}$ with $f(x)gt1$ for all $x in S$. Show that

$$lim_{n rightarrow infty} int_S(f(x))^{frac{1}{n}}=mu(S)$$

I have searched that there are some similar questions here and here, but they do not start from the same hyphothesis.The sequence $f_n = (f(x))^{frac{1}{n}}$ is decreasing and I have tried to solve it by dominated/monotone convergence, but i believe that there is something missing about the integrability of $f$. I do not know how to continue.

nt.number theory – Generalization of $lim_{n rightarrow infty} prod_{i=1}^{n}frac{2n-1}{2n}$ for a character $chi:mathbb{Z}/s mathbb{Z} rightarrow mathbb{C}^*$

Playing with some infinite products I came up with this problem, that I’m not able to figure it out by myself. Moreover in the internet it doesn’t seem to appear anywhere.
Maybe it is just an easy consequence of properties of characters that I’m not aware of, anyway thank you in advance for any help/answers/suggestions.

All of us know (it is fairly easy to see) that $$lim_{n rightarrow infty} frac{1 cdot 3cdot dots cdot (2n-1)}{2 cdot 4 cdot dots cdot (2n)}=0$$

Now this fact could be reformulated in this fashion: let $$chi:mathbb{Z}/2 mathbb{Z} rightarrow mathbb{C}^*$$
the only non trivial character of $mathbb{Z}/2 mathbb{Z}$, then the expression $$lim_{n rightarrow infty}(prod_{n in mathbb{N}}n^{chi(n : text{mod} 2mathbb{Z})})^{-1}=0$$
More generally, we can perform this construction for every $s in mathbb{N}$.

Indeed all we have to do is to consider a non trivial character $$chi:mathbb{Z}/s mathbb{Z} rightarrow mathbb{C}^*$$
and consider the limit
$$lim_{n rightarrow infty}(prod_{n in mathbb{N}}n^{chi(n : text{mod} smathbb{Z})})^{-1}$$

Now it is true that:

  1. The value of the limit is finite for every $s$ and every non trivial character $chi$?
  2. If so the value of the limit depends only on $s$ or also on $chi$?
  3. The limit is always a real number? (Possibly $0$?)