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Continuous function and limit $ f(x,y)=left{begin{matrix} (x-y)sinfrac{1}{x}sinfrac{1}{y} & , xyneq 0 \ 0 &, x=y=0 end{matrix}right. $

2

$begingroup$

I have this function:$$ f(x,y)=left{begin{matrix}
(x-y)sinfrac{1}{x}sinfrac{1}{y} & , xyneq 0 \ 0
&, x=y=0
end{matrix}right. $$
a) Show that $ lim_{xrightarrow 0}[lim_{yrightarrow 0} f(x,y)] $ does not exist.

b) Show that $ f(x,y) $ is continuous at $ (0,0) $ .

For (a) I took $ lim_{yrightarrow 0} f(x,y) $ and I got that it's equal to $$ xsinfrac{1}{x}lim_{yrightarrow 0}sinfrac{1}{y} - sinfrac{1}{x}lim_{yrightarrow 0}y sinfrac{1}{y} $$, but $$ lim_{yrightarrow 0}sinfrac{1}{y} $$ does not exist, so $ lim_{yrightarrow 0} f(x,y) $ does not exist and $ lim_{xrightarrow 0}[lim_{yrightarrow 0} f(x,y)] $ does not exist. Am I right?

For (b) I guess that I have to show that $$ lim_{(x,y)rightarrow (0,0)} f(x,y) = 0 $$ How can I do this ?

limits functions trigonometry continued-fractions

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edited Mar 6 at 0:54

rash

568216

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

$endgroup$

add a comment | 

2

$begingroup$

I have this function:$$ f(x,y)=left{begin{matrix}
(x-y)sinfrac{1}{x}sinfrac{1}{y} & , xyneq 0 \ 0
&, x=y=0
end{matrix}right. $$
a) Show that $ lim_{xrightarrow 0}[lim_{yrightarrow 0} f(x,y)] $ does not exist.

b) Show that $ f(x,y) $ is continuous at $ (0,0) $ .

For (a) I took $ lim_{yrightarrow 0} f(x,y) $ and I got that it's equal to $$ xsinfrac{1}{x}lim_{yrightarrow 0}sinfrac{1}{y} - sinfrac{1}{x}lim_{yrightarrow 0}y sinfrac{1}{y} $$, but $$ lim_{yrightarrow 0}sinfrac{1}{y} $$ does not exist, so $ lim_{yrightarrow 0} f(x,y) $ does not exist and $ lim_{xrightarrow 0}[lim_{yrightarrow 0} f(x,y)] $ does not exist. Am I right?

For (b) I guess that I have to show that $$ lim_{(x,y)rightarrow (0,0)} f(x,y) = 0 $$ How can I do this ?

limits functions trigonometry continued-fractions

share|cite|improve this question

edited Mar 6 at 0:54

rash

568216

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

$endgroup$

add a comment | 

$begingroup$

I have this function:$$ f(x,y)=left{begin{matrix}
(x-y)sinfrac{1}{x}sinfrac{1}{y} & , xyneq 0 \ 0
&, x=y=0
end{matrix}right. $$
a) Show that $ lim_{xrightarrow 0}[lim_{yrightarrow 0} f(x,y)] $ does not exist.

b) Show that $ f(x,y) $ is continuous at $ (0,0) $ .

For (a) I took $ lim_{yrightarrow 0} f(x,y) $ and I got that it's equal to $$ xsinfrac{1}{x}lim_{yrightarrow 0}sinfrac{1}{y} - sinfrac{1}{x}lim_{yrightarrow 0}y sinfrac{1}{y} $$, but $$ lim_{yrightarrow 0}sinfrac{1}{y} $$ does not exist, so $ lim_{yrightarrow 0} f(x,y) $ does not exist and $ lim_{xrightarrow 0}[lim_{yrightarrow 0} f(x,y)] $ does not exist. Am I right?

For (b) I guess that I have to show that $$ lim_{(x,y)rightarrow (0,0)} f(x,y) = 0 $$ How can I do this ?

limits functions trigonometry continued-fractions

share|cite|improve this question

edited Mar 6 at 0:54

rash

568216

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

$endgroup$

share|cite|improve this question

edited Mar 6 at 0:54

rash

568216

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

share|cite|improve this question

edited Mar 6 at 0:54

rash

568216

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

edited Mar 6 at 0:54

rash

568216

edited Mar 6 at 0:54

rash

568216

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

asked Mar 6 at 0:03

Dr.MathematicsDr.Mathematics

516

3 Answers
3

active

oldest

votes

2

$begingroup$

For part (b), I think the easiest proof is just to note that $$|(x-y)sinfrac{1}{x}sinfrac{1}{y}| leq |x-y|.$$

Thus, it's enough to prove that $$lim_{(x,y) to (0,0)}(x-y) = 0.$$

share|cite|improve this answer

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

$endgroup$

add a comment | 

1

$begingroup$

Use polar coordinates $x=rcostheta$ and $y=rsintheta$. The limit is the same as $rto 0$. The original trigonometric functions are always between $-1$ and $1$, $|costheta-sintheta|<2$, so your original expression is always less than $2r$ in absolute value.

share|cite|improve this answer

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For (a), you're not quite where you need to be. You need both that the limit of one term of your difference does not exist and also that the limit of the other term does exist. For example, $lim_{x to 0} (sin (1/x) - sin (1/x)) = 0$ even though neither term of the difference has a limit.
  $endgroup$
  – Robert Shore
  Mar 6 at 0:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I should have specified that this is the answer for part b only. Part a is ok.
  $endgroup$
  – Andrei
  Mar 6 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore yeah I am sorry. In my paper I showed that $ ysinfrac{1}{y}rightarrow 0 $
  $endgroup$
  – Dr.Mathematics
  Mar 6 at 1:00
  

  
  
  
  

add a comment | 

0

$begingroup$

$sin{(1/x)sin(1/y)}$ is an interval and $x-y$ under the limit of $(x,y)to{(0,0)}$ is an interval too. if a function in a close interval is continuous, its valued field is an interval, but its converse proposition may be not true. for example, your function is a counterexample. so you should separate your problem into two cases with $f^{'}(0,0)$ and $lim_{(x,y)to{(0,0)}}f^{'}(x,y)$

the first limit exists but the second do not! for example rewrite your function to $F_{x}^{'}(x,y)=sin(1/x)sin(1/y)+(x-y)sin(1/y)cos(1/x)(-1/x^2)$

$F_{y}^{'}(x,y)=-sin(1/x)sin(1/y)+(x-y)sin(1/x)cos(1/y)(-1/y^2)$

divide both side with $sin(1/x)sin(1/y)$ to get

$1+(x-y)cos(1/x)(-frac{1}{x^{2}sin(1/x)}）$

$-1+(x-y)cos(1/y)(-frac{1}{y^{2}sin(1/y)）}$

since $f(x)=x^2{sin(1/x)}$ or equal to zero exists a limit at $f^{'}(0)$, but $lim_{xto{0}}f^{'}(x)=2xsin(1/x)-cos(1/x)$ do not exists, this result give a certain conclusion to your problem. thank you!

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answered Mar 24 at 10:34

user653679

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2

$begingroup$

For part (b), I think the easiest proof is just to note that $$|(x-y)sinfrac{1}{x}sinfrac{1}{y}| leq |x-y|.$$

Thus, it's enough to prove that $$lim_{(x,y) to (0,0)}(x-y) = 0.$$

share|cite|improve this answer

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

$endgroup$

add a comment | 

2

$begingroup$

For part (b), I think the easiest proof is just to note that $$|(x-y)sinfrac{1}{x}sinfrac{1}{y}| leq |x-y|.$$

Thus, it's enough to prove that $$lim_{(x,y) to (0,0)}(x-y) = 0.$$

share|cite|improve this answer

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

$endgroup$

add a comment | 

$begingroup$

For part (b), I think the easiest proof is just to note that $$|(x-y)sinfrac{1}{x}sinfrac{1}{y}| leq |x-y|.$$

Thus, it's enough to prove that $$lim_{(x,y) to (0,0)}(x-y) = 0.$$

share|cite|improve this answer

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

$endgroup$

share|cite|improve this answer

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

answered Mar 6 at 0:36

Robert ShoreRobert Shore

3,832324

1

$begingroup$

Use polar coordinates $x=rcostheta$ and $y=rsintheta$. The limit is the same as $rto 0$. The original trigonometric functions are always between $-1$ and $1$, $|costheta-sintheta|<2$, so your original expression is always less than $2r$ in absolute value.

share|cite|improve this answer

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For (a), you're not quite where you need to be. You need both that the limit of one term of your difference does not exist and also that the limit of the other term does exist. For example, $lim_{x to 0} (sin (1/x) - sin (1/x)) = 0$ even though neither term of the difference has a limit.
  $endgroup$
  – Robert Shore
  Mar 6 at 0:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I should have specified that this is the answer for part b only. Part a is ok.
  $endgroup$
  – Andrei
  Mar 6 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore yeah I am sorry. In my paper I showed that $ ysinfrac{1}{y}rightarrow 0 $
  $endgroup$
  – Dr.Mathematics
  Mar 6 at 1:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Use polar coordinates $x=rcostheta$ and $y=rsintheta$. The limit is the same as $rto 0$. The original trigonometric functions are always between $-1$ and $1$, $|costheta-sintheta|<2$, so your original expression is always less than $2r$ in absolute value.

share|cite|improve this answer

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For (a), you're not quite where you need to be. You need both that the limit of one term of your difference does not exist and also that the limit of the other term does exist. For example, $lim_{x to 0} (sin (1/x) - sin (1/x)) = 0$ even though neither term of the difference has a limit.
  $endgroup$
  – Robert Shore
  Mar 6 at 0:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, I should have specified that this is the answer for part b only. Part a is ok.
  $endgroup$
  – Andrei
  Mar 6 at 0:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertShore yeah I am sorry. In my paper I showed that $ ysinfrac{1}{y}rightarrow 0 $
  $endgroup$
  – Dr.Mathematics
  Mar 6 at 1:00
  

  
  
  
  

add a comment | 

$begingroup$

Use polar coordinates $x=rcostheta$ and $y=rsintheta$. The limit is the same as $rto 0$. The original trigonometric functions are always between $-1$ and $1$, $|costheta-sintheta|<2$, so your original expression is always less than $2r$ in absolute value.

share|cite|improve this answer

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330

$endgroup$

share|cite|improve this answer

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330

answered Mar 6 at 0:23

AndreiAndrei

13.8k21330