Evaluate $ lim_xrightarrow 0left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right) $ The 2019 Stack Overflow Developer Survey Results Are InCalculate: $lim_n rightarrowinftyleft(frac3^-nsin(3^(1-n))tan(3^1-2n) right)$computing $lim_x to 0 fracsin2x+arctan3x+3x^2lnleft(1+3x+sin^2xright)+xcdot e^x$Solve limit without use L'Hopital: $lim _xto 0left(fracsinleft(2xright)-2sinleft(xright)xcdot :arctg^2xright)$$lim_xto 1/2 left(fractan(pi x)2x-1+frac2pi(2x-1)^2right)$ using L' HopitalShow that $lim_n rightarrow inftyleft(n^2 - frac1sin^2(frac1n) right)=-frac13$Find $lim_xrightarrow fracpi4left(fracsin xcos xright)^left(fracsin 2xcos 2xright)$How can I calculate the following limit? $beginequation* lim_x rightarrow 0 fractan(x) - sin(x)x^3 endequation*,$No L' Hospital $ lim_x rightarrow 0 left (1+frac 1 arctan x right)^sin x$ and $ lim_x rightarrow 0 frac tan ^7 x ln (7x+1) $Calculating $lim_x rightarrow 0 fractan x - sin xx^3$.Evaluate: $lim_xto0left (frac4^tan(x)+ cos(x)2right)^cot(x)$

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Evaluate $ lim_xrightarrow 0left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right) $



The 2019 Stack Overflow Developer Survey Results Are InCalculate: $lim_n rightarrowinftyleft(frac3^-nsin(3^(1-n))tan(3^1-2n) right)$computing $lim_x to 0 fracsin2x+arctan3x+3x^2lnleft(1+3x+sin^2xright)+xcdot e^x$Solve limit without use L'Hopital: $lim _xto 0left(fracsinleft(2xright)-2sinleft(xright)xcdot :arctg^2xright)$$lim_xto 1/2 left(fractan(pi x)2x-1+frac2pi(2x-1)^2right)$ using L' HopitalShow that $lim_n rightarrow inftyleft(n^2 - frac1sin^2(frac1n) right)=-frac13$Find $lim_xrightarrow fracpi4left(fracsin xcos xright)^left(fracsin 2xcos 2xright)$How can I calculate the following limit? $beginequation* lim_x rightarrow 0 fractan(x) - sin(x)x^3 endequation*,$No L' Hospital $ lim_x rightarrow 0 left (1+frac 1 arctan x right)^sin x$ and $ lim_x rightarrow 0 frac tan ^7 x ln (7x+1) $Calculating $lim_x rightarrow 0 fractan x - sin xx^3$.Evaluate: $lim_xto0left (frac4^tan(x)+ cos(x)2right)^cot(x)$










-1












$begingroup$


Calculate the limit
$$
lim_xrightarrow 0left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
$$
My try was to use the Taylor expansion of the denominator and numerator but it wasn't beauty approach.






real-analysis limits







share|cite|improve this question











$endgroup$
















    -1












    $begingroup$


    Calculate the limit
    $$
    lim_xrightarrow 0left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
    $$
    My try was to use the Taylor expansion of the denominator and numerator but it wasn't beauty approach.






    real-analysis limits







    share|cite|improve this question











    $endgroup$














      -1












      -1








      -1


      0



      $begingroup$


      Calculate the limit
      $$
      lim_xrightarrow 0left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
      $$
      My try was to use the Taylor expansion of the denominator and numerator but it wasn't beauty approach.






      real-analysis limits







      share|cite|improve this question











      $endgroup$




      Calculate the limit
      $$
      lim_xrightarrow 0left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
      $$
      My try was to use the Taylor expansion of the denominator and numerator but it wasn't beauty approach.






      real-analysis limits




      real-analysis limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 23 at 16:44









      Jneven

      953322




      953322










      asked Mar 23 at 16:01









      avan1235avan1235

      3598




      3598




















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Let $u=xsin x$



          $$lim_uto 0left(dfrac1utan u-dfrac1u^2right)=lim_uto 0dfracu-tan uu^2tan u$$



          Can you proceed?






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            I've corrected some errors, but the conversion back to $x$-space is unnecessary.
            $endgroup$
            – J.G.
            Mar 23 at 16:13










          • $begingroup$
            I'll remove that then @J.G.Thanks. Cheers :))
            $endgroup$
            – Paras Khosla
            Mar 23 at 16:13










          • $begingroup$
            Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
            $endgroup$
            – avan1235
            Mar 23 at 16:14






          • 2




            $begingroup$
            @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
            $endgroup$
            – J.G.
            Mar 23 at 16:16






          • 1




            $begingroup$
            @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
            $endgroup$
            – Paramanand Singh
            Mar 24 at 3:24


















          2












          $begingroup$

          The function you want to compute the limit of is even, so you can as well consider the limit for $xto0^+$.



          Consider the function $f(x)=xsin x$, defined and positive in a some interval $(0,delta)$, has limit $0$ for $xto0^+$ and $f'(x)=sin x+xcos x$. Thus we can assume $f'(x)>0$ over $(0,delta)$. Thus $f$ is continuous and invertible over $(0,delta)$ and therefore the substitution $u=xsin x$ is possible. Then
          $$
          lim_xto 0^+left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
          =lim_uto0^+fracu-tan uu^2tan u
          $$
          which is easy if you recall that
          $$
          tan u=u+frac13u^3+o(u^3)
          $$






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Let $u=xsin x$



            $$lim_uto 0left(dfrac1utan u-dfrac1u^2right)=lim_uto 0dfracu-tan uu^2tan u$$



            Can you proceed?






            share|cite|improve this answer











            $endgroup$








            • 1




              $begingroup$
              I've corrected some errors, but the conversion back to $x$-space is unnecessary.
              $endgroup$
              – J.G.
              Mar 23 at 16:13










            • $begingroup$
              I'll remove that then @J.G.Thanks. Cheers :))
              $endgroup$
              – Paras Khosla
              Mar 23 at 16:13










            • $begingroup$
              Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
              $endgroup$
              – avan1235
              Mar 23 at 16:14






            • 2




              $begingroup$
              @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
              $endgroup$
              – J.G.
              Mar 23 at 16:16






            • 1




              $begingroup$
              @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
              $endgroup$
              – Paramanand Singh
              Mar 24 at 3:24















            4












            $begingroup$

            Let $u=xsin x$



            $$lim_uto 0left(dfrac1utan u-dfrac1u^2right)=lim_uto 0dfracu-tan uu^2tan u$$



            Can you proceed?






            share|cite|improve this answer











            $endgroup$








            • 1




              $begingroup$
              I've corrected some errors, but the conversion back to $x$-space is unnecessary.
              $endgroup$
              – J.G.
              Mar 23 at 16:13










            • $begingroup$
              I'll remove that then @J.G.Thanks. Cheers :))
              $endgroup$
              – Paras Khosla
              Mar 23 at 16:13










            • $begingroup$
              Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
              $endgroup$
              – avan1235
              Mar 23 at 16:14






            • 2




              $begingroup$
              @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
              $endgroup$
              – J.G.
              Mar 23 at 16:16






            • 1




              $begingroup$
              @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
              $endgroup$
              – Paramanand Singh
              Mar 24 at 3:24













            4












            4








            4





            $begingroup$

            Let $u=xsin x$



            $$lim_uto 0left(dfrac1utan u-dfrac1u^2right)=lim_uto 0dfracu-tan uu^2tan u$$



            Can you proceed?






            share|cite|improve this answer











            $endgroup$



            Let $u=xsin x$



            $$lim_uto 0left(dfrac1utan u-dfrac1u^2right)=lim_uto 0dfracu-tan uu^2tan u$$



            Can you proceed?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 23 at 16:14

























            answered Mar 23 at 16:08









            Paras KhoslaParas Khosla

            3,173626




            3,173626







            • 1




              $begingroup$
              I've corrected some errors, but the conversion back to $x$-space is unnecessary.
              $endgroup$
              – J.G.
              Mar 23 at 16:13










            • $begingroup$
              I'll remove that then @J.G.Thanks. Cheers :))
              $endgroup$
              – Paras Khosla
              Mar 23 at 16:13










            • $begingroup$
              Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
              $endgroup$
              – avan1235
              Mar 23 at 16:14






            • 2




              $begingroup$
              @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
              $endgroup$
              – J.G.
              Mar 23 at 16:16






            • 1




              $begingroup$
              @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
              $endgroup$
              – Paramanand Singh
              Mar 24 at 3:24












            • 1




              $begingroup$
              I've corrected some errors, but the conversion back to $x$-space is unnecessary.
              $endgroup$
              – J.G.
              Mar 23 at 16:13










            • $begingroup$
              I'll remove that then @J.G.Thanks. Cheers :))
              $endgroup$
              – Paras Khosla
              Mar 23 at 16:13










            • $begingroup$
              Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
              $endgroup$
              – avan1235
              Mar 23 at 16:14






            • 2




              $begingroup$
              @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
              $endgroup$
              – J.G.
              Mar 23 at 16:16






            • 1




              $begingroup$
              @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
              $endgroup$
              – Paramanand Singh
              Mar 24 at 3:24







            1




            1




            $begingroup$
            I've corrected some errors, but the conversion back to $x$-space is unnecessary.
            $endgroup$
            – J.G.
            Mar 23 at 16:13




            $begingroup$
            I've corrected some errors, but the conversion back to $x$-space is unnecessary.
            $endgroup$
            – J.G.
            Mar 23 at 16:13












            $begingroup$
            I'll remove that then @J.G.Thanks. Cheers :))
            $endgroup$
            – Paras Khosla
            Mar 23 at 16:13




            $begingroup$
            I'll remove that then @J.G.Thanks. Cheers :))
            $endgroup$
            – Paras Khosla
            Mar 23 at 16:13












            $begingroup$
            Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
            $endgroup$
            – avan1235
            Mar 23 at 16:14




            $begingroup$
            Thanks, whats the official arguemnt to change $xsin(x)$ to $u$?
            $endgroup$
            – avan1235
            Mar 23 at 16:14




            2




            2




            $begingroup$
            @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
            $endgroup$
            – J.G.
            Mar 23 at 16:16




            $begingroup$
            @avan1235 $lim_xto af(g(x))=lim_utolim_xto ag(x)f(u)$ provided $lim_xto ag(x)$ exists.
            $endgroup$
            – J.G.
            Mar 23 at 16:16




            1




            1




            $begingroup$
            @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
            $endgroup$
            – Paramanand Singh
            Mar 24 at 3:24




            $begingroup$
            @avan1235: the rule of substitution in limits is as follows: if $lim_xto a f(x) =L$ and $lim_xto b g(x) =a$ and $g(x) neq a$ as $xto b$ then $$lim_xto b f(g(x)) =L=lim_xto a f(x) $$ Here $g(x)=xsin x, a=b=0$.
            $endgroup$
            – Paramanand Singh
            Mar 24 at 3:24











            2












            $begingroup$

            The function you want to compute the limit of is even, so you can as well consider the limit for $xto0^+$.



            Consider the function $f(x)=xsin x$, defined and positive in a some interval $(0,delta)$, has limit $0$ for $xto0^+$ and $f'(x)=sin x+xcos x$. Thus we can assume $f'(x)>0$ over $(0,delta)$. Thus $f$ is continuous and invertible over $(0,delta)$ and therefore the substitution $u=xsin x$ is possible. Then
            $$
            lim_xto 0^+left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
            =lim_uto0^+fracu-tan uu^2tan u
            $$
            which is easy if you recall that
            $$
            tan u=u+frac13u^3+o(u^3)
            $$






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              The function you want to compute the limit of is even, so you can as well consider the limit for $xto0^+$.



              Consider the function $f(x)=xsin x$, defined and positive in a some interval $(0,delta)$, has limit $0$ for $xto0^+$ and $f'(x)=sin x+xcos x$. Thus we can assume $f'(x)>0$ over $(0,delta)$. Thus $f$ is continuous and invertible over $(0,delta)$ and therefore the substitution $u=xsin x$ is possible. Then
              $$
              lim_xto 0^+left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
              =lim_uto0^+fracu-tan uu^2tan u
              $$
              which is easy if you recall that
              $$
              tan u=u+frac13u^3+o(u^3)
              $$






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                The function you want to compute the limit of is even, so you can as well consider the limit for $xto0^+$.



                Consider the function $f(x)=xsin x$, defined and positive in a some interval $(0,delta)$, has limit $0$ for $xto0^+$ and $f'(x)=sin x+xcos x$. Thus we can assume $f'(x)>0$ over $(0,delta)$. Thus $f$ is continuous and invertible over $(0,delta)$ and therefore the substitution $u=xsin x$ is possible. Then
                $$
                lim_xto 0^+left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
                =lim_uto0^+fracu-tan uu^2tan u
                $$
                which is easy if you recall that
                $$
                tan u=u+frac13u^3+o(u^3)
                $$






                share|cite|improve this answer









                $endgroup$



                The function you want to compute the limit of is even, so you can as well consider the limit for $xto0^+$.



                Consider the function $f(x)=xsin x$, defined and positive in a some interval $(0,delta)$, has limit $0$ for $xto0^+$ and $f'(x)=sin x+xcos x$. Thus we can assume $f'(x)>0$ over $(0,delta)$. Thus $f$ is continuous and invertible over $(0,delta)$ and therefore the substitution $u=xsin x$ is possible. Then
                $$
                lim_xto 0^+left( frac1xsin(x)tan(xsin(x))-frac1x^2sin^2(x) right)
                =lim_uto0^+fracu-tan uu^2tan u
                $$
                which is easy if you recall that
                $$
                tan u=u+frac13u^3+o(u^3)
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 23 at 16:37









                egregegreg

                185k1486208




                185k1486208



























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                    FrenchArabicFrenchAir TamajeqChadian ArabicDazaKaadoTamahaqStandardArabicEnglishDazaDinkaDoniFurKadugliKangaKeigaKrongoTemeinTeseTulishiTumtumZaghawaAcheronAmiraDengebuEbangJomangKatlaKoKwalibLaroLogolLumunMoroNdingNgileShirumbaTagoiTegaliTegemTimaTiroTochoToronaUtoroWarnangBejaDomariSudanese ArabicSudanese Sign LanguagesBertiZaghawa Tebu languagesSaharan languagesLanguages of ChadLanguages of NigerToubou people Nilo-Saharan languageDaza peopleChadGouranDjurab DesertBorkouFaya-LargeauTibesti MountainsChadNigerN'guigmiLibyaSudanOmdurmanJeddahSaudi ArabiaTedaga languageTedaToubou peopleSabhaSaharanKanuri languageKanembu languageTebu languagesZaghawa languageBerti languageFrenchArabicliterary languagebilingualHausaZarmaTuaregromanisation Not to be confused with Dazawa language. "Goran language" redirects here. It is not to be confused with Gorani language (Zaza-Gorani) or Gora dialect. Daza Dazaga Native to Chad, Niger, Sudan, Libya Region Chad, BET Region, bahr el Gazel R...
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