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Examine the uniform continuity $f(x)=fractan(sin x)-frac(sin x)^36x^5$

1

$begingroup$

Let $f: (0, fracpi2)rightarrow mathbb R$, $f(x)=fractan(sin x)-frac(sin x)^36x^5$. Examine the uniform continuity in $(0, fracpi4]$ and in $[fracpi4,fracpi2)$.

I tried to use one of claims:

• 
  $f'$ limited $Rightarrow$ $f$ is uniform continuity
  


• for all $x_n, y_n$ such that $(x_n-y_n) rightarrow 0$ we have $(f(x_n)-f(y_n))rightarrow 0$ then $f$ is uniformly continuous
  


• 
  $forall _epsilon>0 exists_delta>0 forall _x,y in A (|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$ then $f$ is uniformly continuous.
  
  However in this task I have trigonometric functions and I have a problem to use some of this claims and I think that I need some clever way to do it.
  
  Can you get me some tips?

real-analysis

share|cite|improve this question

edited Mar 30 at 22:37

MP3129

asked Mar 30 at 22:16

MP3129MP3129

802211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $lim_xto 0f(x)=infty$, so cannot be uniformly continuous on $(0, fracpi4]$. It is continuous on $[pi/4, pi/2]$ so uniformly continuous on that interval.
  $endgroup$
  – Yu Ding
  Mar 30 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuDing "$lim_x rightarrow 0 f(x)= infty$, so cannot be uniformly continuous on $(0, fracpi4]$" because if $f$ is unlimited then also $f'$ is unlimited, yes?
  $endgroup$
  – MP3129
  Mar 30 at 23:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  for any $delta$, you see $|f(delta')-f(delta)|>1$ for $delta'$ sufficiently close to $0$, so it is not uniformly continuous. I think here one need not think about $f'$, in fact it could be $f'$ not bounded yet $f$ is still uniformly continuous (e.g. $f(x)=xsin frac 1x$). $f'$ being bounded is basically saying $f$ is "Lipschitz".
  $endgroup$
  – Yu Ding
  Mar 30 at 23:07
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $f: (0, fracpi2)rightarrow mathbb R$, $f(x)=fractan(sin x)-frac(sin x)^36x^5$. Examine the uniform continuity in $(0, fracpi4]$ and in $[fracpi4,fracpi2)$.

I tried to use one of claims:

• 
  $f'$ limited $Rightarrow$ $f$ is uniform continuity
  


• for all $x_n, y_n$ such that $(x_n-y_n) rightarrow 0$ we have $(f(x_n)-f(y_n))rightarrow 0$ then $f$ is uniformly continuous
  


• 
  $forall _epsilon>0 exists_delta>0 forall _x,y in A (|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$ then $f$ is uniformly continuous.
  
  However in this task I have trigonometric functions and I have a problem to use some of this claims and I think that I need some clever way to do it.
  
  Can you get me some tips?

real-analysis

share|cite|improve this question

edited Mar 30 at 22:37

MP3129

asked Mar 30 at 22:16

MP3129MP3129

802211

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $lim_xto 0f(x)=infty$, so cannot be uniformly continuous on $(0, fracpi4]$. It is continuous on $[pi/4, pi/2]$ so uniformly continuous on that interval.
  $endgroup$
  – Yu Ding
  Mar 30 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuDing "$lim_x rightarrow 0 f(x)= infty$, so cannot be uniformly continuous on $(0, fracpi4]$" because if $f$ is unlimited then also $f'$ is unlimited, yes?
  $endgroup$
  – MP3129
  Mar 30 at 23:01
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  for any $delta$, you see $|f(delta')-f(delta)|>1$ for $delta'$ sufficiently close to $0$, so it is not uniformly continuous. I think here one need not think about $f'$, in fact it could be $f'$ not bounded yet $f$ is still uniformly continuous (e.g. $f(x)=xsin frac 1x$). $f'$ being bounded is basically saying $f$ is "Lipschitz".
  $endgroup$
  – Yu Ding
  Mar 30 at 23:07
  

  
  
  
  

add a comment | 

$begingroup$

Let $f: (0, fracpi2)rightarrow mathbb R$, $f(x)=fractan(sin x)-frac(sin x)^36x^5$. Examine the uniform continuity in $(0, fracpi4]$ and in $[fracpi4,fracpi2)$.

I tried to use one of claims:

• 
  $f'$ limited $Rightarrow$ $f$ is uniform continuity
  


• for all $x_n, y_n$ such that $(x_n-y_n) rightarrow 0$ we have $(f(x_n)-f(y_n))rightarrow 0$ then $f$ is uniformly continuous
  


• 
  $forall _epsilon>0 exists_delta>0 forall _x,y in A (|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$ then $f$ is uniformly continuous.
  
  However in this task I have trigonometric functions and I have a problem to use some of this claims and I think that I need some clever way to do it.
  
  Can you get me some tips?

real-analysis

share|cite|improve this question

edited Mar 30 at 22:37

MP3129

asked Mar 30 at 22:16

MP3129MP3129

802211

$endgroup$

share|cite|improve this question

edited Mar 30 at 22:37

MP3129

asked Mar 30 at 22:16

MP3129MP3129

802211

share|cite|improve this question

edited Mar 30 at 22:37

MP3129

asked Mar 30 at 22:16

MP3129MP3129

802211

asked Mar 30 at 22:16

MP3129MP3129

802211

asked Mar 30 at 22:16

MP3129MP3129

802211