Can someone help me solve this limit?












0












$begingroup$


Can someone help me solve this limit? Thank you in advance.



$$lim_{xto 0} bigg( frac{1+tan{x}}{1+sin{x}} bigg)^{frac{1}{sin{x}}}$$



If possible without L'Hospital, the exercise gives more points if we solve it without it.






real-analysis calculus limits







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  • 1




    $begingroup$
    Looks like a case for the rule of L'Hospital. Have you tried that?
    $endgroup$
    – Cornman
    3 hours ago










  • $begingroup$
    Hint: $sin x$ and $tan x$ can be replaced by $x$, to the first order. The limit is that of $(1+o(x))^{1/x}$.
    $endgroup$
    – Yves Daoust
    3 hours ago












  • $begingroup$
    I have forgot to add the last part that if it is possible to not use L'Hospital , because if we use it in places where they think we could have used something else , they penalise us
    $endgroup$
    – user644728
    3 hours ago










  • $begingroup$
    "Can someone help me solve this limit?" And how is that supposed to help you learn?
    $endgroup$
    – John Coleman
    2 hours ago
















0












$begingroup$


Can someone help me solve this limit? Thank you in advance.



$$lim_{xto 0} bigg( frac{1+tan{x}}{1+sin{x}} bigg)^{frac{1}{sin{x}}}$$



If possible without L'Hospital, the exercise gives more points if we solve it without it.






real-analysis calculus limits







share|cite|improve this question









New contributor




user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Looks like a case for the rule of L'Hospital. Have you tried that?
    $endgroup$
    – Cornman
    3 hours ago










  • $begingroup$
    Hint: $sin x$ and $tan x$ can be replaced by $x$, to the first order. The limit is that of $(1+o(x))^{1/x}$.
    $endgroup$
    – Yves Daoust
    3 hours ago












  • $begingroup$
    I have forgot to add the last part that if it is possible to not use L'Hospital , because if we use it in places where they think we could have used something else , they penalise us
    $endgroup$
    – user644728
    3 hours ago










  • $begingroup$
    "Can someone help me solve this limit?" And how is that supposed to help you learn?
    $endgroup$
    – John Coleman
    2 hours ago














0












0








0





$begingroup$


Can someone help me solve this limit? Thank you in advance.



$$lim_{xto 0} bigg( frac{1+tan{x}}{1+sin{x}} bigg)^{frac{1}{sin{x}}}$$



If possible without L'Hospital, the exercise gives more points if we solve it without it.






real-analysis calculus limits







share|cite|improve this question









New contributor




user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Can someone help me solve this limit? Thank you in advance.



$$lim_{xto 0} bigg( frac{1+tan{x}}{1+sin{x}} bigg)^{frac{1}{sin{x}}}$$



If possible without L'Hospital, the exercise gives more points if we solve it without it.






real-analysis calculus limits




real-analysis calculus limits






share|cite|improve this question









New contributor




user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Lemniscate

402211




402211






New contributor




user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









user644728user644728

325




325




New contributor




user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user644728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Looks like a case for the rule of L'Hospital. Have you tried that?
    $endgroup$
    – Cornman
    3 hours ago










  • $begingroup$
    Hint: $sin x$ and $tan x$ can be replaced by $x$, to the first order. The limit is that of $(1+o(x))^{1/x}$.
    $endgroup$
    – Yves Daoust
    3 hours ago












  • $begingroup$
    I have forgot to add the last part that if it is possible to not use L'Hospital , because if we use it in places where they think we could have used something else , they penalise us
    $endgroup$
    – user644728
    3 hours ago










  • $begingroup$
    "Can someone help me solve this limit?" And how is that supposed to help you learn?
    $endgroup$
    – John Coleman
    2 hours ago














  • 1




    $begingroup$
    Looks like a case for the rule of L'Hospital. Have you tried that?
    $endgroup$
    – Cornman
    3 hours ago










  • $begingroup$
    Hint: $sin x$ and $tan x$ can be replaced by $x$, to the first order. The limit is that of $(1+o(x))^{1/x}$.
    $endgroup$
    – Yves Daoust
    3 hours ago












  • $begingroup$
    I have forgot to add the last part that if it is possible to not use L'Hospital , because if we use it in places where they think we could have used something else , they penalise us
    $endgroup$
    – user644728
    3 hours ago










  • $begingroup$
    "Can someone help me solve this limit?" And how is that supposed to help you learn?
    $endgroup$
    – John Coleman
    2 hours ago








1




1




$begingroup$
Looks like a case for the rule of L'Hospital. Have you tried that?
$endgroup$
– Cornman
3 hours ago




$begingroup$
Looks like a case for the rule of L'Hospital. Have you tried that?
$endgroup$
– Cornman
3 hours ago












$begingroup$
Hint: $sin x$ and $tan x$ can be replaced by $x$, to the first order. The limit is that of $(1+o(x))^{1/x}$.
$endgroup$
– Yves Daoust
3 hours ago






$begingroup$
Hint: $sin x$ and $tan x$ can be replaced by $x$, to the first order. The limit is that of $(1+o(x))^{1/x}$.
$endgroup$
– Yves Daoust
3 hours ago














$begingroup$
I have forgot to add the last part that if it is possible to not use L'Hospital , because if we use it in places where they think we could have used something else , they penalise us
$endgroup$
– user644728
3 hours ago




$begingroup$
I have forgot to add the last part that if it is possible to not use L'Hospital , because if we use it in places where they think we could have used something else , they penalise us
$endgroup$
– user644728
3 hours ago












$begingroup$
"Can someone help me solve this limit?" And how is that supposed to help you learn?
$endgroup$
– John Coleman
2 hours ago




$begingroup$
"Can someone help me solve this limit?" And how is that supposed to help you learn?
$endgroup$
– John Coleman
2 hours ago










3 Answers
3






active

oldest

votes


















1












$begingroup$

$hspace{5mm} lim_{xto 0}left(frac{1+tan(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



$=lim_{xto 0}left(1+frac{1+tan(x)}{1+sin(x)}-1right)^{tfrac{1}{sin(x)}}$



$=lim_{xto 0}left(1+frac{tan(x)-sin(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



$=lim_{xto 0}left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{tfrac{1}{sin(x)}}$



$=lim_{xto 0}left[left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{frac{1+sin(x)}{tan(x)-sin(x)}}right]^{tfrac{1}{sin(x)} cdot frac{tan(x)-sin(x)}{1+sin(x)}}$



$=mathrm{exp} big( {lim_{xto 0}tfrac{1}{sin(x)} frac{tan(x)-sin(x)}{1+sin(x)}} big)$



$=mathrm{exp} big( {lim_{xto 0}frac{1-cos(x)}{cos(x)(1+sin(x))}} big)$



$=mathrm{exp} big( {frac{1-1}{1(1+0)}} big)$



$=e^0$



$=1$






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    The searched limit is the ratio



    $$frac{lim_{xto0}(1+tan x)^{1/sin x}}{lim_{xto0}(1+sin x)^{1/sin x}}=frac{lim_{xto0}((1+tan x)^{1/tan x})^{1/cos x}}{lim_{xto0}(1+sin x)^{1/sin x}}.$$



    As $tan x$ and $sin x$ tend to zero continuously with $x$, $$frac{e^1}e.$$






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      Let
      $$L=lim_{xto 0} (frac{1+tan{(x)}}{1+sin{(x)}})^{frac{1}{sin{(x)}}}$$
      $$ln{L}=lim_{xto 0} frac{1}{sin{(x)}}ln{(frac{1+tan{(x)}}{1+sin{(x)}})}$$
      $$ln{L}=lim_{xto 0} frac{ln{(1+tan{(x)})}-ln{(1+sin{(x)})})}{sin{(x)}}$$
      $$ln{L}=lim_{xto 0} frac{frac{sec^2{(x)}}{1+tan{(x)}}-frac{cos{(x)}}{1+sin{(x)}}}{cos{(x)}}$$
      $$ln{L}=lim_{xto 0} frac{1}{cos^3{(x)}(1+tan{(x)})}-frac{1}{1+sin{(x)}}$$
      $$ln{L}= 1-1=0$$
      $$therefore L=e^0=1$$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        $hspace{5mm} lim_{xto 0}left(frac{1+tan(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



        $=lim_{xto 0}left(1+frac{1+tan(x)}{1+sin(x)}-1right)^{tfrac{1}{sin(x)}}$



        $=lim_{xto 0}left(1+frac{tan(x)-sin(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



        $=lim_{xto 0}left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{tfrac{1}{sin(x)}}$



        $=lim_{xto 0}left[left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{frac{1+sin(x)}{tan(x)-sin(x)}}right]^{tfrac{1}{sin(x)} cdot frac{tan(x)-sin(x)}{1+sin(x)}}$



        $=mathrm{exp} big( {lim_{xto 0}tfrac{1}{sin(x)} frac{tan(x)-sin(x)}{1+sin(x)}} big)$



        $=mathrm{exp} big( {lim_{xto 0}frac{1-cos(x)}{cos(x)(1+sin(x))}} big)$



        $=mathrm{exp} big( {frac{1-1}{1(1+0)}} big)$



        $=e^0$



        $=1$






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

          $hspace{5mm} lim_{xto 0}left(frac{1+tan(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



          $=lim_{xto 0}left(1+frac{1+tan(x)}{1+sin(x)}-1right)^{tfrac{1}{sin(x)}}$



          $=lim_{xto 0}left(1+frac{tan(x)-sin(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



          $=lim_{xto 0}left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{tfrac{1}{sin(x)}}$



          $=lim_{xto 0}left[left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{frac{1+sin(x)}{tan(x)-sin(x)}}right]^{tfrac{1}{sin(x)} cdot frac{tan(x)-sin(x)}{1+sin(x)}}$



          $=mathrm{exp} big( {lim_{xto 0}tfrac{1}{sin(x)} frac{tan(x)-sin(x)}{1+sin(x)}} big)$



          $=mathrm{exp} big( {lim_{xto 0}frac{1-cos(x)}{cos(x)(1+sin(x))}} big)$



          $=mathrm{exp} big( {frac{1-1}{1(1+0)}} big)$



          $=e^0$



          $=1$






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

            $hspace{5mm} lim_{xto 0}left(frac{1+tan(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left(1+frac{1+tan(x)}{1+sin(x)}-1right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left(1+frac{tan(x)-sin(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left[left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{frac{1+sin(x)}{tan(x)-sin(x)}}right]^{tfrac{1}{sin(x)} cdot frac{tan(x)-sin(x)}{1+sin(x)}}$



            $=mathrm{exp} big( {lim_{xto 0}tfrac{1}{sin(x)} frac{tan(x)-sin(x)}{1+sin(x)}} big)$



            $=mathrm{exp} big( {lim_{xto 0}frac{1-cos(x)}{cos(x)(1+sin(x))}} big)$



            $=mathrm{exp} big( {frac{1-1}{1(1+0)}} big)$



            $=e^0$



            $=1$






            share|cite|improve this answer











            $endgroup$



            $hspace{5mm} lim_{xto 0}left(frac{1+tan(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left(1+frac{1+tan(x)}{1+sin(x)}-1right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left(1+frac{tan(x)-sin(x)}{1+sin(x)}right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{tfrac{1}{sin(x)}}$



            $=lim_{xto 0}left[left(1+frac{1}{big(frac{1+sin(x)}{tan(x)-sin(x)}big)}right)^{frac{1+sin(x)}{tan(x)-sin(x)}}right]^{tfrac{1}{sin(x)} cdot frac{tan(x)-sin(x)}{1+sin(x)}}$



            $=mathrm{exp} big( {lim_{xto 0}tfrac{1}{sin(x)} frac{tan(x)-sin(x)}{1+sin(x)}} big)$



            $=mathrm{exp} big( {lim_{xto 0}frac{1-cos(x)}{cos(x)(1+sin(x))}} big)$



            $=mathrm{exp} big( {frac{1-1}{1(1+0)}} big)$



            $=e^0$



            $=1$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago









            Lemniscate

            402211




            402211










            answered 3 hours ago









            user326159user326159

            1,2521722




            1,2521722























                4












                $begingroup$

                The searched limit is the ratio



                $$frac{lim_{xto0}(1+tan x)^{1/sin x}}{lim_{xto0}(1+sin x)^{1/sin x}}=frac{lim_{xto0}((1+tan x)^{1/tan x})^{1/cos x}}{lim_{xto0}(1+sin x)^{1/sin x}}.$$



                As $tan x$ and $sin x$ tend to zero continuously with $x$, $$frac{e^1}e.$$






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  The searched limit is the ratio



                  $$frac{lim_{xto0}(1+tan x)^{1/sin x}}{lim_{xto0}(1+sin x)^{1/sin x}}=frac{lim_{xto0}((1+tan x)^{1/tan x})^{1/cos x}}{lim_{xto0}(1+sin x)^{1/sin x}}.$$



                  As $tan x$ and $sin x$ tend to zero continuously with $x$, $$frac{e^1}e.$$






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    The searched limit is the ratio



                    $$frac{lim_{xto0}(1+tan x)^{1/sin x}}{lim_{xto0}(1+sin x)^{1/sin x}}=frac{lim_{xto0}((1+tan x)^{1/tan x})^{1/cos x}}{lim_{xto0}(1+sin x)^{1/sin x}}.$$



                    As $tan x$ and $sin x$ tend to zero continuously with $x$, $$frac{e^1}e.$$






                    share|cite|improve this answer











                    $endgroup$



                    The searched limit is the ratio



                    $$frac{lim_{xto0}(1+tan x)^{1/sin x}}{lim_{xto0}(1+sin x)^{1/sin x}}=frac{lim_{xto0}((1+tan x)^{1/tan x})^{1/cos x}}{lim_{xto0}(1+sin x)^{1/sin x}}.$$



                    As $tan x$ and $sin x$ tend to zero continuously with $x$, $$frac{e^1}e.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 hours ago

























                    answered 3 hours ago









                    Yves DaoustYves Daoust

                    128k674227




                    128k674227























                        2












                        $begingroup$

                        Let
                        $$L=lim_{xto 0} (frac{1+tan{(x)}}{1+sin{(x)}})^{frac{1}{sin{(x)}}}$$
                        $$ln{L}=lim_{xto 0} frac{1}{sin{(x)}}ln{(frac{1+tan{(x)}}{1+sin{(x)}})}$$
                        $$ln{L}=lim_{xto 0} frac{ln{(1+tan{(x)})}-ln{(1+sin{(x)})})}{sin{(x)}}$$
                        $$ln{L}=lim_{xto 0} frac{frac{sec^2{(x)}}{1+tan{(x)}}-frac{cos{(x)}}{1+sin{(x)}}}{cos{(x)}}$$
                        $$ln{L}=lim_{xto 0} frac{1}{cos^3{(x)}(1+tan{(x)})}-frac{1}{1+sin{(x)}}$$
                        $$ln{L}= 1-1=0$$
                        $$therefore L=e^0=1$$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Let
                          $$L=lim_{xto 0} (frac{1+tan{(x)}}{1+sin{(x)}})^{frac{1}{sin{(x)}}}$$
                          $$ln{L}=lim_{xto 0} frac{1}{sin{(x)}}ln{(frac{1+tan{(x)}}{1+sin{(x)}})}$$
                          $$ln{L}=lim_{xto 0} frac{ln{(1+tan{(x)})}-ln{(1+sin{(x)})})}{sin{(x)}}$$
                          $$ln{L}=lim_{xto 0} frac{frac{sec^2{(x)}}{1+tan{(x)}}-frac{cos{(x)}}{1+sin{(x)}}}{cos{(x)}}$$
                          $$ln{L}=lim_{xto 0} frac{1}{cos^3{(x)}(1+tan{(x)})}-frac{1}{1+sin{(x)}}$$
                          $$ln{L}= 1-1=0$$
                          $$therefore L=e^0=1$$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Let
                            $$L=lim_{xto 0} (frac{1+tan{(x)}}{1+sin{(x)}})^{frac{1}{sin{(x)}}}$$
                            $$ln{L}=lim_{xto 0} frac{1}{sin{(x)}}ln{(frac{1+tan{(x)}}{1+sin{(x)}})}$$
                            $$ln{L}=lim_{xto 0} frac{ln{(1+tan{(x)})}-ln{(1+sin{(x)})})}{sin{(x)}}$$
                            $$ln{L}=lim_{xto 0} frac{frac{sec^2{(x)}}{1+tan{(x)}}-frac{cos{(x)}}{1+sin{(x)}}}{cos{(x)}}$$
                            $$ln{L}=lim_{xto 0} frac{1}{cos^3{(x)}(1+tan{(x)})}-frac{1}{1+sin{(x)}}$$
                            $$ln{L}= 1-1=0$$
                            $$therefore L=e^0=1$$






                            share|cite|improve this answer









                            $endgroup$



                            Let
                            $$L=lim_{xto 0} (frac{1+tan{(x)}}{1+sin{(x)}})^{frac{1}{sin{(x)}}}$$
                            $$ln{L}=lim_{xto 0} frac{1}{sin{(x)}}ln{(frac{1+tan{(x)}}{1+sin{(x)}})}$$
                            $$ln{L}=lim_{xto 0} frac{ln{(1+tan{(x)})}-ln{(1+sin{(x)})})}{sin{(x)}}$$
                            $$ln{L}=lim_{xto 0} frac{frac{sec^2{(x)}}{1+tan{(x)}}-frac{cos{(x)}}{1+sin{(x)}}}{cos{(x)}}$$
                            $$ln{L}=lim_{xto 0} frac{1}{cos^3{(x)}(1+tan{(x)})}-frac{1}{1+sin{(x)}}$$
                            $$ln{L}= 1-1=0$$
                            $$therefore L=e^0=1$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            Peter ForemanPeter Foreman

                            2,13613




                            2,13613






















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                                Hivernacle

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                                Hivernacle Hivernacle a les cases del Salt a Alcoi Un hivernacle és una construcció més o menys permanent que permet modificar el microclima del seu interior i permetre el conreu de vegetals en millors condicions o fora d'època. [1] Són el nivell superior de l'anomenat conreu protegit que va des d'un mínim que correspon a l'ús de plàstics que cobreixen el terra (mulching) fins al màxim que serien els hivernacles de vidre i amb un sistema complet de calefacció. Etimològicament la paraula hivernacle es refereix al lloc on passen l'hivern les plantes. En la pràctica actual s'hi mantenen tota la temporada de conreu o més. Contingut 1 Funcionament 2 Evolució en l'ús dels hivernacles 3 Característiques constructives dels hivernacles 4 Hivernacles als Països Baixos 5 Referències Funcionament Els que no tenen calefacció aprofiten precisament l'efecte hivernacle que produeix un escalfame
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                                Fluorita

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                                Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac
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                                Hulsita

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                                Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà
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