Evaluating $lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$












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$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$






calculus limits radicals limits-without-lhopital







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  • $begingroup$
    Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
    $endgroup$
    – coreyman317
    Jan 14 at 16:40












  • $begingroup$
    As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
    $endgroup$
    – Turan Nasibli
    Jan 14 at 16:44










  • $begingroup$
    While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
    $endgroup$
    – JimB
    Jan 14 at 17:17










  • $begingroup$
    Just because they are asymptotically equivalent does not mean they are simply equal.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 17:18










  • $begingroup$
    @JimB I know the answer as well,thanks anyway.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 17:18
















0












$begingroup$


$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$






calculus limits radicals limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
    $endgroup$
    – coreyman317
    Jan 14 at 16:40












  • $begingroup$
    As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
    $endgroup$
    – Turan Nasibli
    Jan 14 at 16:44










  • $begingroup$
    While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
    $endgroup$
    – JimB
    Jan 14 at 17:17










  • $begingroup$
    Just because they are asymptotically equivalent does not mean they are simply equal.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 17:18










  • $begingroup$
    @JimB I know the answer as well,thanks anyway.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 17:18














0












0








0





$begingroup$


$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$






calculus limits radicals limits-without-lhopital







share|cite|improve this question











$endgroup$




$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$






calculus limits radicals limits-without-lhopital




calculus limits radicals limits-without-lhopital






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edited Jan 14 at 18:56









Michael Rozenberg

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asked Jan 14 at 16:36









Turan NasibliTuran Nasibli

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846












  • $begingroup$
    Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
    $endgroup$
    – coreyman317
    Jan 14 at 16:40












  • $begingroup$
    As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
    $endgroup$
    – Turan Nasibli
    Jan 14 at 16:44










  • $begingroup$
    While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
    $endgroup$
    – JimB
    Jan 14 at 17:17










  • $begingroup$
    Just because they are asymptotically equivalent does not mean they are simply equal.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 17:18










  • $begingroup$
    @JimB I know the answer as well,thanks anyway.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 17:18


















  • $begingroup$
    Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
    $endgroup$
    – coreyman317
    Jan 14 at 16:40












  • $begingroup$
    As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
    $endgroup$
    – Turan Nasibli
    Jan 14 at 16:44










  • $begingroup$
    While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
    $endgroup$
    – JimB
    Jan 14 at 17:17










  • $begingroup$
    Just because they are asymptotically equivalent does not mean they are simply equal.
    $endgroup$
    – Simply Beautiful Art
    Jan 14 at 17:18










  • $begingroup$
    @JimB I know the answer as well,thanks anyway.
    $endgroup$
    – Turan Nasibli
    Jan 14 at 17:18
















$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40






$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40














$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nasibli
Jan 14 at 16:44




$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nasibli
Jan 14 at 16:44












$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17




$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17












$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18




$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18












$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nasibli
Jan 14 at 17:18




$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nasibli
Jan 14 at 17:18










2 Answers
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$begingroup$

$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






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    $begingroup$

    $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



    Set $x=1/h^2,hto0^+$



    $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



    $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



    $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



    $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






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      2 Answers
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      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      2












      $begingroup$

      $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
      $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
      $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
      $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
        $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
        $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
        $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
          $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
          $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
          $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






          share|cite|improve this answer









          $endgroup$



          $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
          $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
          $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
          $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 18:54









          Michael RozenbergMichael Rozenberg

          111k1897201




          111k1897201























              0












              $begingroup$

              $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



              Set $x=1/h^2,hto0^+$



              $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



              $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



              $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



              $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



                Set $x=1/h^2,hto0^+$



                $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



                $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



                $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



                $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



                  Set $x=1/h^2,hto0^+$



                  $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



                  $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



                  $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



                  $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






                  share|cite|improve this answer









                  $endgroup$



                  $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



                  Set $x=1/h^2,hto0^+$



                  $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



                  $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



                  $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



                  $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$







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                  answered Jan 17 at 14:52









                  lab bhattacharjeelab bhattacharjee

                  229k15159279




                  229k15159279






























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                      Måne

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                      För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
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                      Storängen

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                      För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
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                      VLT Carioca

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                      VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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