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Lower limit of a real function

Real Analysis Monotone Convergence Theorem QuestionProve that $liminf x_n = -limsup (-x_n)$$X_nleq Y_n$ implies $liminf X_n leq liminf Y_n$ and $limsup X_n leq limsup Y_n$Show that if the sequence$(x_n)$ is bounded, then $(x_n)$ converges iff $limsup(x_n)=liminf(x_n)$.Limit of the sequence given by $x_n+1=x_n-x_n^n+1$If $a_n$ is bounded sequence , prove that $limsuplimits_nto infty a_n=-liminflimits_nto infty(-a_n)$Understanding the definitions of limit superior and limit inferior of a real sequenceLimit of square root of sequence given value of limit of original sequenceRequest for assistance on finishing the proof that $lim_ntoinftyn x_n = 0$ given some initial conditions.Upper Limit Definition

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$begingroup$

We define $f:X to Bbb R$ as a function, and $g: (0,1]to Bbb R$ defined by $$g(r) := inf_xin Br(x_0)f(x).$$
Denote $$
liminf_xto x_0f(x) := lim_rto 0^+g(r) = lim_rto 0^+inf_x∈Br(x_0)f(x).$$

i) Show that if $x_nto x_0$, then $liminflimits_nto inftyf(x_n) geq liminflimits_xto x_0 f(x)$.

ii) Show that there exists a sequence $x_n$ converging to $x_0$ such that $limlimits_nto inftyf(x_n) = liminflimits_xto x_0f(x).$

I have proven that limit of $g(r)$ as $r$ approaches $0$ exists and it is monotone decreasing. But I am confused as in isn't the definition of $liminflimits_xto x_0=inf liminflimits_nto infty f(x_n)$, which would prove both i) and ii) as trivial?

real-analysis sequences-and-series limits

share|cite|improve this question

edited 21 hours ago

DanielWainfleet

35.7k31648

asked yesterday

james blackjames black

419113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My edit was for a minor typo.
  $endgroup$
  – DanielWainfleet
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: The last line. We have $lim inf_xto x_0f(x)=inf lim inf_nto inftyf(x_n): (x_n)_nin Bbb Nin S$ where $S$ is the set of $ all $ sequences that converge to $x_0$. There may be some $(x_n)_xin Bbb Nin S$ such that $lim inf_nto infty f(x_n)$ does not converge to $lim inf_xto x_0f(x)$. For example let $f(x)=1$ when $xin Bbb Q$ and $f(x)=0$ when $xnot in Bbb Q,$ and $x_0=0.$ Then $f(1/n)to 1$ as $nto infty$ but $lim inf_xto 0f(x)=0.$
  $endgroup$
  – DanielWainfleet
  21 hours ago
  

  
  
  
  

add a comment | 

0

1

$begingroup$

We define $f:X to Bbb R$ as a function, and $g: (0,1]to Bbb R$ defined by $$g(r) := inf_xin Br(x_0)f(x).$$
Denote $$
liminf_xto x_0f(x) := lim_rto 0^+g(r) = lim_rto 0^+inf_x∈Br(x_0)f(x).$$

i) Show that if $x_nto x_0$, then $liminflimits_nto inftyf(x_n) geq liminflimits_xto x_0 f(x)$.

ii) Show that there exists a sequence $x_n$ converging to $x_0$ such that $limlimits_nto inftyf(x_n) = liminflimits_xto x_0f(x).$

I have proven that limit of $g(r)$ as $r$ approaches $0$ exists and it is monotone decreasing. But I am confused as in isn't the definition of $liminflimits_xto x_0=inf liminflimits_nto infty f(x_n)$, which would prove both i) and ii) as trivial?

real-analysis sequences-and-series limits

share|cite|improve this question

edited 21 hours ago

DanielWainfleet

35.7k31648

asked yesterday

james blackjames black

419113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My edit was for a minor typo.
  $endgroup$
  – DanielWainfleet
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Re: The last line. We have $lim inf_xto x_0f(x)=inf lim inf_nto inftyf(x_n): (x_n)_nin Bbb Nin S$ where $S$ is the set of $ all $ sequences that converge to $x_0$. There may be some $(x_n)_xin Bbb Nin S$ such that $lim inf_nto infty f(x_n)$ does not converge to $lim inf_xto x_0f(x)$. For example let $f(x)=1$ when $xin Bbb Q$ and $f(x)=0$ when $xnot in Bbb Q,$ and $x_0=0.$ Then $f(1/n)to 1$ as $nto infty$ but $lim inf_xto 0f(x)=0.$
  $endgroup$
  – DanielWainfleet
  21 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We define $f:X to Bbb R$ as a function, and $g: (0,1]to Bbb R$ defined by $$g(r) := inf_xin Br(x_0)f(x).$$
Denote $$
liminf_xto x_0f(x) := lim_rto 0^+g(r) = lim_rto 0^+inf_x∈Br(x_0)f(x).$$

i) Show that if $x_nto x_0$, then $liminflimits_nto inftyf(x_n) geq liminflimits_xto x_0 f(x)$.

ii) Show that there exists a sequence $x_n$ converging to $x_0$ such that $limlimits_nto inftyf(x_n) = liminflimits_xto x_0f(x).$

I have proven that limit of $g(r)$ as $r$ approaches $0$ exists and it is monotone decreasing. But I am confused as in isn't the definition of $liminflimits_xto x_0=inf liminflimits_nto infty f(x_n)$, which would prove both i) and ii) as trivial?

real-analysis sequences-and-series limits

share|cite|improve this question

edited 21 hours ago

DanielWainfleet

35.7k31648

asked yesterday

james blackjames black

419113

$endgroup$

share|cite|improve this question

edited 21 hours ago

DanielWainfleet

35.7k31648

asked yesterday

james blackjames black

419113

share|cite|improve this question

edited 21 hours ago

DanielWainfleet

35.7k31648

asked yesterday

james blackjames black

419113

edited 21 hours ago

DanielWainfleet

35.7k31648

edited 21 hours ago

DanielWainfleet

35.7k31648

asked yesterday

james blackjames black

419113

asked yesterday

james blackjames black

419113