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
    9 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
    9 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
    8 hours ago










  • $begingroup$
    "Can someone help me solve this limit?" And how is that supposed to help you learn?
    $endgroup$
    – John Coleman
    8 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
    9 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
    9 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
    8 hours ago










  • $begingroup$
    "Can someone help me solve this limit?" And how is that supposed to help you learn?
    $endgroup$
    – John Coleman
    8 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 8 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 9 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
    9 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
    9 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
    8 hours ago










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














  • 1




    $begingroup$
    Looks like a case for the rule of L'Hospital. Have you tried that?
    $endgroup$
    – Cornman
    9 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
    9 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
    8 hours ago










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








1




1




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




$begingroup$
Looks like a case for the rule of L'Hospital. Have you tried that?
$endgroup$
– Cornman
9 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
9 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
9 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
8 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
8 hours ago












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




$begingroup$
"Can someone help me solve this limit?" And how is that supposed to help you learn?
$endgroup$
– John Coleman
8 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











$endgroup$





















    5












    $begingroup$

    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 8 hours ago









            Lemniscate

            402211




            402211










            answered 8 hours ago









            user326159user326159

            1,2621722




            1,2621722























                5












                $begingroup$

                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











                $endgroup$


















                  5












                  $begingroup$

                  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











                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    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 8 hours ago

























                    answered 8 hours ago









                    Yves DaoustYves Daoust

                    128k675227




                    128k675227























                        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 8 hours ago









                            Peter ForemanPeter Foreman

                            2,13613




                            2,13613






















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                                How to label and detect the document text images

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                                1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
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                                Vallis Paradisi

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                                Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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                                Tabula Rosettana

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                                Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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