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

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





















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









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071874%2fis-it-appropriate-to-say-that-lim-x-to-4-sqrt-x-2-is-both-0-and-4%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071874%2fis-it-appropriate-to-say-that-lim-x-to-4-sqrt-x-2-is-both-0-and-4%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .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
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
            Read more

            Gyllenstierna

            Image
            .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
            Read more