The lift and the right hand sides for Piecewise limit












2












$begingroup$


Considering this limit :



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$.



from the lift :



$limlimits_{x to 1-} (x+1) = 2 $



from the right:



$limlimits_{x to 1+} (pi) = pi $



I'm assuming that this limit is not exist since the left hand side limit does not equal the right hand side limit (From my current knowledge).



The Book that i use telling me not what i expected !:



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$ = $limlimits_{x to 1} (x+1) = 2 $



I think i miss some things about Piecewise limits. anyone explain to me why this limit end up with 2 ?






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are not calculating the limit from the right correctly. This limit should also be $2$.
    $endgroup$
    – platty
    Dec 1 '18 at 9:05
















2












$begingroup$


Considering this limit :



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$.



from the lift :



$limlimits_{x to 1-} (x+1) = 2 $



from the right:



$limlimits_{x to 1+} (pi) = pi $



I'm assuming that this limit is not exist since the left hand side limit does not equal the right hand side limit (From my current knowledge).



The Book that i use telling me not what i expected !:



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$ = $limlimits_{x to 1} (x+1) = 2 $



I think i miss some things about Piecewise limits. anyone explain to me why this limit end up with 2 ?






calculus limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are not calculating the limit from the right correctly. This limit should also be $2$.
    $endgroup$
    – platty
    Dec 1 '18 at 9:05














2












2








2





$begingroup$


Considering this limit :



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$.



from the lift :



$limlimits_{x to 1-} (x+1) = 2 $



from the right:



$limlimits_{x to 1+} (pi) = pi $



I'm assuming that this limit is not exist since the left hand side limit does not equal the right hand side limit (From my current knowledge).



The Book that i use telling me not what i expected !:



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$ = $limlimits_{x to 1} (x+1) = 2 $



I think i miss some things about Piecewise limits. anyone explain to me why this limit end up with 2 ?






calculus limits







share|cite|improve this question











$endgroup$




Considering this limit :



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$.



from the lift :



$limlimits_{x to 1-} (x+1) = 2 $



from the right:



$limlimits_{x to 1+} (pi) = pi $



I'm assuming that this limit is not exist since the left hand side limit does not equal the right hand side limit (From my current knowledge).



The Book that i use telling me not what i expected !:



$limlimits_{x to 1} =
begin{cases}
x+1, & text{x≠1} \[2ex]
pi , & text{x=1}
end{cases}$ = $limlimits_{x to 1} (x+1) = 2 $



I think i miss some things about Piecewise limits. anyone explain to me why this limit end up with 2 ?






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 10:27







Ammar Bamhdi

















asked Dec 1 '18 at 9:03









Ammar BamhdiAmmar Bamhdi

204




204








  • 1




    $begingroup$
    You are not calculating the limit from the right correctly. This limit should also be $2$.
    $endgroup$
    – platty
    Dec 1 '18 at 9:05














  • 1




    $begingroup$
    You are not calculating the limit from the right correctly. This limit should also be $2$.
    $endgroup$
    – platty
    Dec 1 '18 at 9:05








1




1




$begingroup$
You are not calculating the limit from the right correctly. This limit should also be $2$.
$endgroup$
– platty
Dec 1 '18 at 9:05




$begingroup$
You are not calculating the limit from the right correctly. This limit should also be $2$.
$endgroup$
– platty
Dec 1 '18 at 9:05










1 Answer
1






active

oldest

votes


















4












$begingroup$

Recall that, according to the definition, when we take the limit $xto 1$ we are assuming $xneq 1$, that is



$$forall varepsilon>0 quad exists delta>0 quad text{such that}quad color{green}{forall xneq1}quad|x-1|<delta implies|f(x)-2|<varepsilon$$



therefore since $xneq 1$



$$limlimits_{x to 1} f(x)=limlimits_{x to 1} (x+1) = 2$$



In other words, the value for the limit at a point is not affected by the value of the function at that point. The function could be also not defined at that point (e.g. $sin x/x$ as $x to 0$).



Your evaluation would be correct for the following function



$$g(x)=
begin{cases}
x+1, & text{$x<1$} \[2ex]
pi & text{$xge1$}
end{cases}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    OK, got the idea
    $endgroup$
    – Ammar Bamhdi
    Dec 1 '18 at 9:27










  • $begingroup$
    @AmmarBamhdi Well done! Keep always that in mind. Bye
    $endgroup$
    – gimusi
    Dec 1 '18 at 9:28











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1 Answer
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1 Answer
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active

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active

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active

oldest

votes









4












$begingroup$

Recall that, according to the definition, when we take the limit $xto 1$ we are assuming $xneq 1$, that is



$$forall varepsilon>0 quad exists delta>0 quad text{such that}quad color{green}{forall xneq1}quad|x-1|<delta implies|f(x)-2|<varepsilon$$



therefore since $xneq 1$



$$limlimits_{x to 1} f(x)=limlimits_{x to 1} (x+1) = 2$$



In other words, the value for the limit at a point is not affected by the value of the function at that point. The function could be also not defined at that point (e.g. $sin x/x$ as $x to 0$).



Your evaluation would be correct for the following function



$$g(x)=
begin{cases}
x+1, & text{$x<1$} \[2ex]
pi & text{$xge1$}
end{cases}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    OK, got the idea
    $endgroup$
    – Ammar Bamhdi
    Dec 1 '18 at 9:27










  • $begingroup$
    @AmmarBamhdi Well done! Keep always that in mind. Bye
    $endgroup$
    – gimusi
    Dec 1 '18 at 9:28
















4












$begingroup$

Recall that, according to the definition, when we take the limit $xto 1$ we are assuming $xneq 1$, that is



$$forall varepsilon>0 quad exists delta>0 quad text{such that}quad color{green}{forall xneq1}quad|x-1|<delta implies|f(x)-2|<varepsilon$$



therefore since $xneq 1$



$$limlimits_{x to 1} f(x)=limlimits_{x to 1} (x+1) = 2$$



In other words, the value for the limit at a point is not affected by the value of the function at that point. The function could be also not defined at that point (e.g. $sin x/x$ as $x to 0$).



Your evaluation would be correct for the following function



$$g(x)=
begin{cases}
x+1, & text{$x<1$} \[2ex]
pi & text{$xge1$}
end{cases}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    OK, got the idea
    $endgroup$
    – Ammar Bamhdi
    Dec 1 '18 at 9:27










  • $begingroup$
    @AmmarBamhdi Well done! Keep always that in mind. Bye
    $endgroup$
    – gimusi
    Dec 1 '18 at 9:28














4












4








4





$begingroup$

Recall that, according to the definition, when we take the limit $xto 1$ we are assuming $xneq 1$, that is



$$forall varepsilon>0 quad exists delta>0 quad text{such that}quad color{green}{forall xneq1}quad|x-1|<delta implies|f(x)-2|<varepsilon$$



therefore since $xneq 1$



$$limlimits_{x to 1} f(x)=limlimits_{x to 1} (x+1) = 2$$



In other words, the value for the limit at a point is not affected by the value of the function at that point. The function could be also not defined at that point (e.g. $sin x/x$ as $x to 0$).



Your evaluation would be correct for the following function



$$g(x)=
begin{cases}
x+1, & text{$x<1$} \[2ex]
pi & text{$xge1$}
end{cases}$$






share|cite|improve this answer











$endgroup$



Recall that, according to the definition, when we take the limit $xto 1$ we are assuming $xneq 1$, that is



$$forall varepsilon>0 quad exists delta>0 quad text{such that}quad color{green}{forall xneq1}quad|x-1|<delta implies|f(x)-2|<varepsilon$$



therefore since $xneq 1$



$$limlimits_{x to 1} f(x)=limlimits_{x to 1} (x+1) = 2$$



In other words, the value for the limit at a point is not affected by the value of the function at that point. The function could be also not defined at that point (e.g. $sin x/x$ as $x to 0$).



Your evaluation would be correct for the following function



$$g(x)=
begin{cases}
x+1, & text{$x<1$} \[2ex]
pi & text{$xge1$}
end{cases}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 9:17

























answered Dec 1 '18 at 9:04









gimusigimusi

1




1












  • $begingroup$
    OK, got the idea
    $endgroup$
    – Ammar Bamhdi
    Dec 1 '18 at 9:27










  • $begingroup$
    @AmmarBamhdi Well done! Keep always that in mind. Bye
    $endgroup$
    – gimusi
    Dec 1 '18 at 9:28


















  • $begingroup$
    OK, got the idea
    $endgroup$
    – Ammar Bamhdi
    Dec 1 '18 at 9:27










  • $begingroup$
    @AmmarBamhdi Well done! Keep always that in mind. Bye
    $endgroup$
    – gimusi
    Dec 1 '18 at 9:28
















$begingroup$
OK, got the idea
$endgroup$
– Ammar Bamhdi
Dec 1 '18 at 9:27




$begingroup$
OK, got the idea
$endgroup$
– Ammar Bamhdi
Dec 1 '18 at 9:27












$begingroup$
@AmmarBamhdi Well done! Keep always that in mind. Bye
$endgroup$
– gimusi
Dec 1 '18 at 9:28




$begingroup$
@AmmarBamhdi Well done! Keep always that in mind. Bye
$endgroup$
– gimusi
Dec 1 '18 at 9:28


















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