Limit of a quotient.

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Let $X$ a metric space, $E subseteq X$ and $p$ be a limit point of $E$. Let $f,g: E subseteq X to mathbbC$ be functions. In Rudin and Apostol's books, I read the following:

Let $$fracfg: Z:=x in E: g(x) neq 0 to mathbbC: x mapsto fracf(x)g(x)$$

Let $A = lim_x to p f(x), B = lim_x to p g(x)$

Then $$lim_x to p left(fracfgright)(x) = fracAB$$

How are we sure that the limit makes sense? I.e., how do we know that
$p$ is a limit point of $Z$ as well?

real-analysis limits

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asked Jul 26 at 7:33

Math_QED

6,34831344

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up vote
0
down vote

favorite

Let $X$ a metric space, $E subseteq X$ and $p$ be a limit point of $E$. Let $f,g: E subseteq X to mathbbC$ be functions. In Rudin and Apostol's books, I read the following:

Let $$fracfg: Z:=x in E: g(x) neq 0 to mathbbC: x mapsto fracf(x)g(x)$$

Let $A = lim_x to p f(x), B = lim_x to p g(x)$

Then $$lim_x to p left(fracfgright)(x) = fracAB$$

How are we sure that the limit makes sense? I.e., how do we know that
$p$ is a limit point of $Z$ as well?

real-analysis limits

share|cite|improve this question

asked Jul 26 at 7:33

Math_QED

6,34831344

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $X$ a metric space, $E subseteq X$ and $p$ be a limit point of $E$. Let $f,g: E subseteq X to mathbbC$ be functions. In Rudin and Apostol's books, I read the following:

Let $$fracfg: Z:=x in E: g(x) neq 0 to mathbbC: x mapsto fracf(x)g(x)$$

Let $A = lim_x to p f(x), B = lim_x to p g(x)$

Then $$lim_x to p left(fracfgright)(x) = fracAB$$

How are we sure that the limit makes sense? I.e., how do we know that
$p$ is a limit point of $Z$ as well?

real-analysis limits

share|cite|improve this question

asked Jul 26 at 7:33

Math_QED

6,34831344

Let $X$ a metric space, $E subseteq X$ and $p$ be a limit point of $E$. Let $f,g: E subseteq X to mathbbC$ be functions. In Rudin and Apostol's books, I read the following:

Let $$fracfg: Z:=x in E: g(x) neq 0 to mathbbC: x mapsto fracf(x)g(x)$$

Let $A = lim_x to p f(x), B = lim_x to p g(x)$

Then $$lim_x to p left(fracfgright)(x) = fracAB$$

How are we sure that the limit makes sense? I.e., how do we know that
$p$ is a limit point of $Z$ as well?

real-analysis limits

share|cite|improve this question

asked Jul 26 at 7:33

Math_QED

6,34831344

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 7:33

Math_QED

6,34831344

asked Jul 26 at 7:33

Math_QED

6,34831344

asked Jul 26 at 7:33

Math_QED

6,34831344

6,34831344

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

We have to require that $B = lim_x to p g(x) ne 0$.

If this is the case, then $p$ is a limit point of $Z$.

share|cite|improve this answer

answered Jul 26 at 7:38

Fred

37.2k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why is a limit point of $Z$ then?
  – Math_QED
  Jul 26 at 7:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
  – Fred
  Jul 26 at 7:50
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

We have to require that $B = lim_x to p g(x) ne 0$.

If this is the case, then $p$ is a limit point of $Z$.

share|cite|improve this answer

answered Jul 26 at 7:38

Fred

37.2k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why is a limit point of $Z$ then?
  – Math_QED
  Jul 26 at 7:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
  – Fred
  Jul 26 at 7:50
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

We have to require that $B = lim_x to p g(x) ne 0$.

If this is the case, then $p$ is a limit point of $Z$.

share|cite|improve this answer

answered Jul 26 at 7:38

Fred

37.2k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why is a limit point of $Z$ then?
  – Math_QED
  Jul 26 at 7:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
  – Fred
  Jul 26 at 7:50
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

We have to require that $B = lim_x to p g(x) ne 0$.

If this is the case, then $p$ is a limit point of $Z$.

share|cite|improve this answer

answered Jul 26 at 7:38

Fred

37.2k1237

We have to require that $B = lim_x to p g(x) ne 0$.

If this is the case, then $p$ is a limit point of $Z$.

share|cite|improve this answer

answered Jul 26 at 7:38

Fred

37.2k1237

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 7:38

Fred

37.2k1237

answered Jul 26 at 7:38

Fred

37.2k1237

answered Jul 26 at 7:38

Fred

37.2k1237

37.2k1237

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why is a limit point of $Z$ then?
  – Math_QED
  Jul 26 at 7:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
  – Fred
  Jul 26 at 7:50
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  And why is a limit point of $Z$ then?
  – Math_QED
  Jul 26 at 7:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
  – Fred
  Jul 26 at 7:50
  
  

  
  
  
  

And why is a limit point of $Z$ then?
– Math_QED
Jul 26 at 7:46

And why is a limit point of $Z$ then?
– Math_QED
Jul 26 at 7:46

1

1

If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
– Fred
Jul 26 at 7:50

If $B ne 0$, then there is a neighborhood $U$ of $p$ such that $g(x) ne 0$ for all $x in E cap(U setminus p)$.
– Fred
Jul 26 at 7:50

add a comment | 

 

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