Is it appropriate to say that $lim_{x to 4}{(sqrt x-2)}$ is both $0$ and $-4$?












1












$begingroup$


Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?



Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.



Is it correct? if so, why?



Thank you.






limits roots radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sqrt{x^2} = |x|$ is in the definition of square root.
    $endgroup$
    – OmG
    Jan 13 at 10:41










  • $begingroup$
    no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
    $endgroup$
    – Yanko
    Jan 13 at 10:44








  • 4




    $begingroup$
    No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
    $endgroup$
    – KM101
    Jan 13 at 10:48












  • $begingroup$
    @KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
    $endgroup$
    – user477343
    Jan 13 at 11:17








  • 3




    $begingroup$
    It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
    $endgroup$
    – Blue
    Jan 13 at 11:21
















1












$begingroup$


Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?



Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.



Is it correct? if so, why?



Thank you.






limits roots radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sqrt{x^2} = |x|$ is in the definition of square root.
    $endgroup$
    – OmG
    Jan 13 at 10:41










  • $begingroup$
    no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
    $endgroup$
    – Yanko
    Jan 13 at 10:44








  • 4




    $begingroup$
    No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
    $endgroup$
    – KM101
    Jan 13 at 10:48












  • $begingroup$
    @KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
    $endgroup$
    – user477343
    Jan 13 at 11:17








  • 3




    $begingroup$
    It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
    $endgroup$
    – Blue
    Jan 13 at 11:21














1












1








1


1



$begingroup$


Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?



Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.



Is it correct? if so, why?



Thank you.






limits roots radicals







share|cite|improve this question











$endgroup$




Our teachers seem to ignore that, but if I take a limit of a function with square root, for example:
$$lim_{x to 4}{(sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?



Because in the solutions they gave us, nobody mentions the fact that $sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.



Is it correct? if so, why?



Thank you.






limits roots radicals




limits roots radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 11:13









Blue

49.7k870158




49.7k870158










asked Jan 13 at 10:39









NetanelNetanel

1075




1075












  • $begingroup$
    $sqrt{x^2} = |x|$ is in the definition of square root.
    $endgroup$
    – OmG
    Jan 13 at 10:41










  • $begingroup$
    no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
    $endgroup$
    – Yanko
    Jan 13 at 10:44








  • 4




    $begingroup$
    No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
    $endgroup$
    – KM101
    Jan 13 at 10:48












  • $begingroup$
    @KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
    $endgroup$
    – user477343
    Jan 13 at 11:17








  • 3




    $begingroup$
    It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
    $endgroup$
    – Blue
    Jan 13 at 11:21


















  • $begingroup$
    $sqrt{x^2} = |x|$ is in the definition of square root.
    $endgroup$
    – OmG
    Jan 13 at 10:41










  • $begingroup$
    no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
    $endgroup$
    – Yanko
    Jan 13 at 10:44








  • 4




    $begingroup$
    No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
    $endgroup$
    – KM101
    Jan 13 at 10:48












  • $begingroup$
    @KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
    $endgroup$
    – user477343
    Jan 13 at 11:17








  • 3




    $begingroup$
    It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
    $endgroup$
    – Blue
    Jan 13 at 11:21
















$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41




$begingroup$
$sqrt{x^2} = |x|$ is in the definition of square root.
$endgroup$
– OmG
Jan 13 at 10:41












$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44






$begingroup$
no a limit of real numbers can only have one value. Note that $sqrt{x}>0$ so there's no way you get a limit that is less than $-2$.
$endgroup$
– Yanko
Jan 13 at 10:44






4




4




$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48






$begingroup$
No, by convention/definition, the square root function always returns a non-negative value, so $sqrt{x} = vert xvert geq 0$. Here, $sqrt{4} = +2$ only.
$endgroup$
– KM101
Jan 13 at 10:48














$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17






$begingroup$
@KM101 I don't mean to be picky here, but I think you mean $sqrt{x^{color{red}{2}}}=|x|geq0$ for which you did not write the $2$, but we all know what you mean ;)
$endgroup$
– user477343
Jan 13 at 11:17






3




3




$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21




$begingroup$
It's true that $x^2=4$ has two solutions: $x=pm 2$. (More generally, $x^2=m$ has two solutions $x=pm sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $sqrt{4}$ has two values, $pm 2$. When we write "$sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.)
$endgroup$
– Blue
Jan 13 at 11:21










2 Answers
2






active

oldest

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3












$begingroup$

This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.



In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.



However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.




Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
$$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
    $endgroup$
    – manooooh
    Mar 14 at 21:00





















4












$begingroup$

According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.



    In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.



    However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.




    Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
    $$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
    and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
      $endgroup$
      – manooooh
      Mar 14 at 21:00


















    3












    $begingroup$

    This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.



    In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.



    However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.




    Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
    $$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
    and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
      $endgroup$
      – manooooh
      Mar 14 at 21:00
















    3












    3








    3





    $begingroup$

    This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.



    In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.



    However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.




    Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
    $$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
    and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.






    share|cite|improve this answer











    $endgroup$



    This is a common misconception about the function $sqrt{cdot},:[0,infty)toBbb R$ as the "inverse" of $f(x)=x^2$.



    In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.



    However, we can restrict its domain to $[0,infty)$ and the function $flvert_{[0,infty)}$ would be injective there, hence we can define $$sqrt{x}:=flvert_{[0,infty)}^{-1}.$$ The function $sqrt{x}$ thus takes only one positive value for each input $xge 0$ (e.g. $sqrt{4}=2$) even though there are two numbers such that $x^2=4$.




    Remark: There are some cases where $sqrt{cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $sqrt{cdot}$ wouldn't be a "function" in that case. Note also that we could very well define
    $$sqrt{x}:=flvert_{(-infty,0]}^{-1}$$
    and we'd have $sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 13 at 13:19

























    answered Jan 13 at 11:53









    BigbearZzzBigbearZzz

    9,07421753




    9,07421753












    • $begingroup$
      I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
      $endgroup$
      – manooooh
      Mar 14 at 21:00




















    • $begingroup$
      I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
      $endgroup$
      – manooooh
      Mar 14 at 21:00


















    $begingroup$
    I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
    $endgroup$
    – manooooh
    Mar 14 at 21:00






    $begingroup$
    I would also say "but $f(x)=x^2$ is not injective (e.g. $f(2)=f(−2)=4$ but $pmb{2neq-2}$)".
    $endgroup$
    – manooooh
    Mar 14 at 21:00













    4












    $begingroup$

    According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.






        share|cite|improve this answer









        $endgroup$



        According to definition of principal square root this is impossible. The principal square root of $xge0$ is denoted by $sqrt x$ and is always non-negative.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 11:57









        Mostafa AyazMostafa Ayaz

        18.1k31040




        18.1k31040






























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            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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