Did I make a mistake when finding the intervals this function is continous on?












3












$begingroup$


I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!






continuity piecewise-continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, and I am giving the answer in terms of $a$.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 17:59










  • $begingroup$
    You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
    $endgroup$
    – caverac
    Jan 1 at 18:04










  • $begingroup$
    Why is this not continuous at $a$?
    $endgroup$
    – Mark Bennet
    Jan 1 at 18:04










  • $begingroup$
    @MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:06












  • $begingroup$
    @caverac, thanks :) I can't accept your answer since you only posted a comment.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:07
















3












$begingroup$


I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!






continuity piecewise-continuity







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, and I am giving the answer in terms of $a$.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 17:59










  • $begingroup$
    You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
    $endgroup$
    – caverac
    Jan 1 at 18:04










  • $begingroup$
    Why is this not continuous at $a$?
    $endgroup$
    – Mark Bennet
    Jan 1 at 18:04










  • $begingroup$
    @MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:06












  • $begingroup$
    @caverac, thanks :) I can't accept your answer since you only posted a comment.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:07














3












3








3





$begingroup$


I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!






continuity piecewise-continuity







share|cite|improve this question









$endgroup$




I was given a function f given by $f(x) =
begin{cases}
frac{x^2 - a^2}{x - a ;} textit{if} ; x neq a, \
2a ; textit{if} ; x=a \
end{cases}$,
and told to find the intervals over which $f$ is continous, in terms of $a$. I just figured it is continous everywhere except $a$, so the intervals would be $]-infty, a[$ and $]a, infty[$. However, my textbook gives the answer, without working, as $]-infty, -a[$ and $]-a, infty[$. I'm like 99% sure this is a mistake by the textbook author, not me, but continuty and the like is not my strong side, so I just want to make sure I didn't make a mistake here. Thanks in advance!






continuity piecewise-continuity




continuity piecewise-continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 1 at 17:55









Christoffer Corfield AakreChristoffer Corfield Aakre

284




284








  • 1




    $begingroup$
    Yes, and I am giving the answer in terms of $a$.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 17:59










  • $begingroup$
    You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
    $endgroup$
    – caverac
    Jan 1 at 18:04










  • $begingroup$
    Why is this not continuous at $a$?
    $endgroup$
    – Mark Bennet
    Jan 1 at 18:04










  • $begingroup$
    @MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:06












  • $begingroup$
    @caverac, thanks :) I can't accept your answer since you only posted a comment.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:07














  • 1




    $begingroup$
    Yes, and I am giving the answer in terms of $a$.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 17:59










  • $begingroup$
    You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
    $endgroup$
    – caverac
    Jan 1 at 18:04










  • $begingroup$
    Why is this not continuous at $a$?
    $endgroup$
    – Mark Bennet
    Jan 1 at 18:04










  • $begingroup$
    @MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:06












  • $begingroup$
    @caverac, thanks :) I can't accept your answer since you only posted a comment.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:07








1




1




$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59




$begingroup$
Yes, and I am giving the answer in terms of $a$.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 17:59












$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04




$begingroup$
You are right, the function is continuos everywhere except at $x = a$ ... Yet another textbook error :)
$endgroup$
– caverac
Jan 1 at 18:04












$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04




$begingroup$
Why is this not continuous at $a$?
$endgroup$
– Mark Bennet
Jan 1 at 18:04












$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06






$begingroup$
@MarkBennet it is not continous because $limlimits_{x to a}f(x)$ is undefined.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:06














$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07




$begingroup$
@caverac, thanks :) I can't accept your answer since you only posted a comment.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:07










1 Answer
1






active

oldest

votes


















4












$begingroup$

It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:14








  • 1




    $begingroup$
    I think the given solution is wrong(in your textbook)
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 1 at 18:18










  • $begingroup$
    Yeah I figured. Thanks a lot.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:21










  • $begingroup$
    To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
    $endgroup$
    – DanielWainfleet
    Jan 1 at 18:43











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:14








  • 1




    $begingroup$
    I think the given solution is wrong(in your textbook)
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 1 at 18:18










  • $begingroup$
    Yeah I figured. Thanks a lot.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:21










  • $begingroup$
    To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
    $endgroup$
    – DanielWainfleet
    Jan 1 at 18:43
















4












$begingroup$

It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:14








  • 1




    $begingroup$
    I think the given solution is wrong(in your textbook)
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 1 at 18:18










  • $begingroup$
    Yeah I figured. Thanks a lot.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:21










  • $begingroup$
    To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
    $endgroup$
    – DanielWainfleet
    Jan 1 at 18:43














4












4








4





$begingroup$

It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$






share|cite|improve this answer









$endgroup$



It is $$frac{x^2-a^2}{x-a}=x+a$$ if $xneq$ $a$ and
$$lim_{xto a}frac{x^2-a^2}{x-a}=2a$$ thus our function is continous for all real $x$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 1 at 18:10









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.8k42866




76.8k42866












  • $begingroup$
    I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:14








  • 1




    $begingroup$
    I think the given solution is wrong(in your textbook)
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 1 at 18:18










  • $begingroup$
    Yeah I figured. Thanks a lot.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:21










  • $begingroup$
    To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
    $endgroup$
    – DanielWainfleet
    Jan 1 at 18:43


















  • $begingroup$
    I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:14








  • 1




    $begingroup$
    I think the given solution is wrong(in your textbook)
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 1 at 18:18










  • $begingroup$
    Yeah I figured. Thanks a lot.
    $endgroup$
    – Christoffer Corfield Aakre
    Jan 1 at 18:21










  • $begingroup$
    To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
    $endgroup$
    – DanielWainfleet
    Jan 1 at 18:43
















$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14






$begingroup$
I think this is the right answer. I just made a mistake earlier when evaluating $limlimits_{x to a}f(x)$, but now I agree it is contionus for all $xinmathbb{R}$
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:14






1




1




$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18




$begingroup$
I think the given solution is wrong(in your textbook)
$endgroup$
– Dr. Sonnhard Graubner
Jan 1 at 18:18












$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21




$begingroup$
Yeah I figured. Thanks a lot.
$endgroup$
– Christoffer Corfield Aakre
Jan 1 at 18:21












$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43




$begingroup$
To the proposer: $f(x)=x+a$ for all $x$. Clearly this is continuous for all $x.$
$endgroup$
– DanielWainfleet
Jan 1 at 18:43


















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