How to calculate the two limits?












3












$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits







share|cite











$endgroup$












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















3












$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits







share|cite











$endgroup$












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago














3












3








3





$begingroup$



I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits







share|cite











$endgroup$





I got stuck on two exercises below
$$
limlimits_{xrightarrow +infty} left(frac{2}{pi} arctan x right)^x \
lim_{xrightarrow 3^+} frac{cos x ln(x-3)}{ln(e^x-e^3)}
$$




For the first one , let $y=(frac{2}{pi} arctan x )^x $, so $ln y =xln (frac{2}{pi} arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $frac{infty}{infty}$ or $frac{0}{0}$. But when I use the L 'hopital's rule to the $frac{infty}{infty}$ or $frac{0}{0}$ the calculation is complex and useless.



For the second one , it is $frac{infty}{infty}$ type, also useless the L 'hopital's rule is. How to calculate it ?






limits




limits






share|cite















share|cite













share|cite




share|cite








edited 1 hour ago







lanse7pty

















asked 2 hours ago









lanse7ptylanse7pty

1,8411823




1,8411823












  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago


















  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago




$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac{0}{0}$ or $frac{infty}{infty}$, where you can try using L'Hopital's (though it may not help): just remember that $ab = frac{a}{ frac{1}{b} }$. And, L'Hopital's Rule is applicable for $frac{infty}{infty}$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago










4 Answers
4






active

oldest

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2












$begingroup$

Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Without L'Hospital
    $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



    Now, by Taylor for large values of $x$
    $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
    $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
    $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
    $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        You can solve the first one using




        • $arctan x + operatorname{arccot}x = frac{pi}{2}$

        • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

        • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


        begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
        & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
        & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
        end{eqnarray*}



        The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





        • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






        share|cite|improve this answer









        $endgroup$














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          4 Answers
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          4 Answers
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          active

          oldest

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          2












          $begingroup$

          Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$






              share|cite|improve this answer











              $endgroup$



              Rewrite $inftycdot 0$ as $infty cdot dfrac{1}{infty}$. Now you can apply L'Hopital's rule: $$lim_{xto +infty}dfrac{left(ln 2/picdotarctan x right)}{1/x}=lim_{xto +infty}dfrac{pi/2cdot arctan x}{-1/x^2}cdot dfrac{1}{1+x^2}=-dfrac{pi }{2}lim_{xto +infty}arctan xcdot dfrac{x^2}{1+x^2}$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Paras KhoslaParas Khosla

              2,736423




              2,736423























                  1












                  $begingroup$

                  Without L'Hospital
                  $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                  Now, by Taylor for large values of $x$
                  $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                  $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                  $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                  $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Without L'Hospital
                    $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                    Now, by Taylor for large values of $x$
                    $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                    $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                    $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                    $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Without L'Hospital
                      $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                      $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                      $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                      $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached






                      share|cite|improve this answer









                      $endgroup$



                      Without L'Hospital
                      $$y=left(frac{2}{pi} arctan (x) right)^ximplies log(y)=x logleft(frac{2}{pi} arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=frac{pi }{2}-frac{1}{x}+frac{1}{3 x^3}+Oleft(frac{1}{x^4}right)$$
                      $$frac{2}{pi} arctan (x) =1-frac{2}{pi x}+frac{2}{3 pi x^3}+Oleft(frac{1}{x^4}right)$$ Taylor again
                      $$logleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi x}-frac{2}{pi ^2 x^2}+Oleft(frac{1}{x^3}right)$$
                      $$log(y)=xlogleft(frac{2}{pi} arctan (x) right)= -frac{2}{pi }-frac{2}{pi ^2 x}+Oleft(frac{1}{x^2}right)$$ Just continue with Taylor using $y=e^{log(y)}$ if you want to see not only the limit but also how it is approached







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Claude LeiboviciClaude Leibovici

                      125k1158136




                      125k1158136























                          0












                          $begingroup$

                          I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.






                              share|cite|improve this answer









                              $endgroup$



                              I believe you can apply L'hopital's rule for an indeterminate form like $frac{infty}{infty}$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              AdmuthAdmuth

                              385




                              385























                                  0












                                  $begingroup$

                                  You can solve the first one using




                                  • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                  • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                  • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                  begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                  & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                  & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                  end{eqnarray*}



                                  The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                  • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You can solve the first one using




                                    • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                    • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                    • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                    begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                    & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                    & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                    end{eqnarray*}



                                    The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                    • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You can solve the first one using




                                      • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                      • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                      • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                      begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                      & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                      & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                      end{eqnarray*}



                                      The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                      • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.






                                      share|cite|improve this answer









                                      $endgroup$



                                      You can solve the first one using




                                      • $arctan x + operatorname{arccot}x = frac{pi}{2}$

                                      • $lim_{yto 0}(1-y)^{1/y} = e^{-1}$

                                      • $xoperatorname{arccot}x stackrel{stackrel{x =cot u}{uto 0^+}}{=} cot ucdot u = cos ucdot frac{u}{sin u} stackrel{u to 0^+}{longrightarrow} 1$


                                      begin{eqnarray*} left(frac{2}{pi} arctan x right)^x
                                      & stackrel{arctan x = frac{pi}{2}-operatorname{arccot}x}{=} & left( underbrace{left(1- frac{2}{pi}operatorname{arccot}xright)^{frac{pi}{2operatorname{arccot}x}}}_{stackrel{x to +infty}{longrightarrow} e^{-1}} right)^{frac{2}{pi}underbrace{xoperatorname{arccot}x}_{stackrel{x to +infty}{longrightarrow} 1}} \
                                      & stackrel{x to +infty}{longrightarrow} & e^{-frac{2}{pi}}
                                      end{eqnarray*}



                                      The second limit is quite straight forward as $lim_{xto 3+}cos x = cos 3$. Just consider





                                      • $frac{ln(x-3)}{ln(e^x-e^3)}$ and apply L'Hospital.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 11 mins ago









                                      trancelocationtrancelocation

                                      13.4k1827




                                      13.4k1827






























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                                          Усть-Каменогорск

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                                          У этого термина существуют и другие значения, см. Усть-Каменогорск. Город Усть-Каменогорск каз. Өскемен Флаг Герб 49°57′ с. ш. 82°37′ в. д. H G Я O L Страна Казахстан   Казахстан Статус областной центр Область Восточно-Казахстанская Аким Жаксылык Омар [1] История и география Основан 1720 Город с 1868 Площадь 303,15 [2] км² Высота центра 283 ± 1 м Тип климата резко-континентальный Часовой пояс UTC+6 Население Население 329 090 [3]  человек ( 2018 ) Национальности русские 53,56 % казахи 42,99 % немцы 0,82 % украинцы 0,57 % татары 0,79 % белорусы 0,14 % азербайджанцы 0,23 % корейцы 0,20 % прочие 1,08 % -1 шт [4] Этнохороним усть-каменогорцы, усть-каменогорец Цифровые идентификаторы Телефонный код +7 (7232) Почтовый индекс F0***** [5] Автомобильный код 16 (ранее F) oskemen.kz Усть-Каменогорск Усть-Каменогорск ...
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                                          Халкинская богословская школа

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                                          Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης Главное здание Халкинской школы Международное название Theological School of Halki Год основания 1844 Конфессия православие Церковь Константинопольская Ректор Елпидифор (Ламбриниадис) Расположение о. Хейбелиада, Турция Юридический адрес Ümit tepesi No 5, 34973 Heybeliada, Istanbul, Turkey Сайт www.greece.org  [[commons:Category:Halki seminary|Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης на Викискладе]] Ха́лкинская богосло́вская шко́ла (или Духовная семинария на Халки , тур. Rum Ortodoks Ruhban Okulu , греч. Ιερά Θεολογική Σχολή της Χάλκης ) — недействующее среднее специальное учебное заведение Константинопольской православной церкви, расположенное на территории Свято-Троицкого мужского монастыря на вершине холма Надежды (тур. Ümit Tepesi ) острова Хейбелиада, в Турции....
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                                          Высокополье (Харьковская область)

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                                          У этого термина существуют и другие значения, см. Высокополье. Село Высокополье укр. Високопілля Герб 49°55′07″ с. ш. 35°25′13″ в. д. H G Я O L Страна Украина   Украина Область Харьковская Район Валковский Сельский совет Высокопольский История и география Основан 1670 Прежние названия Хмелевое Площадь 55,56 км² Часовой пояс UTC+2, летом UTC+3 Население Население 1957 человек ( 2001 ) Плотность 35,220 чел./км² Цифровые идентификаторы Телефонный код +380 5753 Почтовый индекс 63020 Автомобильный код AX, КХ / 21 КОАТУУ 6321281501 Высокополье Высокополье Высокополье  Аудио, фото и видео на Викискладе Высокопо́лье (укр. Високопілля ) — село, Высокопольский сельский совет, Валковский район, Харьковская область, Украина. Код КОАТУУ — 6321281501. Население по переписи 2001 г. составляет 1957 (866/1091 м/ж) человек. Является административным центр...
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