Limit of $lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$ (No L'Hôpital)












5















$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$




I can't get to the end of this limit. Here is what I worked out:



begin{align*}
& lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
= lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
= & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
end{align*}



Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)






calculus limits trigonometry limits-without-lhopital







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    5















    $lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$




    I can't get to the end of this limit. Here is what I worked out:



    begin{align*}
    & lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
    lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
    = lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
    = & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
    = lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
    = & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
    = lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
    = & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
    = lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
    = & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
    end{align*}



    Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)






    calculus limits trigonometry limits-without-lhopital







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      5












      5








      5








      $lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$




      I can't get to the end of this limit. Here is what I worked out:



      begin{align*}
      & lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      end{align*}



      Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)






      calculus limits trigonometry limits-without-lhopital







      share|cite|improve this question
















      $lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))$




      I can't get to the end of this limit. Here is what I worked out:



      begin{align*}
      & lim_{x to 0} frac{cos 2x}{sin 2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      lim_{x to 0}frac{frac{cos2x }{2x}}{frac{sin 2x}{2x}}cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} frac{cos 2x}{2x} cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} frac{{cos^2 (x)}-{sin^2 (x)}}{2x}cdotfrac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} left(frac{cos^2(x)}{2x}-frac{sin^2 x}{2x}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} left(frac{1-sin^2 x}{2x}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} left(frac{1}{2x}-frac{sin^2 x}{2x}-frac{sin x}{2}right) cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} left(frac{1}{2x}-frac{sin x}{2}-frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      = lim_{x to 0} left(frac{1}{2x}-2frac{sin x}{2}right)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )} \
      = & lim_{x to 0} left(frac{1}{2x}-sin xright)cdot frac{cos(frac{pi }{2}-x )}{sin(frac{pi }{2}-x )}
      end{align*}



      Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)






      calculus limits trigonometry limits-without-lhopital




      calculus limits trigonometry limits-without-lhopital






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      edited Nov 24 at 18:57









      quid♦

      36.9k95093




      36.9k95093










      asked Nov 24 at 13:14









      Jakcjones

      318




      318






















          5 Answers
          5






          active

          oldest

          votes


















          4














          HINT



          We have that



          $$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$






          share|cite|improve this answer























          • Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:27










          • Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
            – gimusi
            Nov 24 at 13:42



















          3














          Hint:
          $$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$






          share|cite|improve this answer





















          • Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:28



















          3














          $displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$



          $displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $






          share|cite|improve this answer





















          • Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:27



















          2














          Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
          $$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$






          share|cite|improve this answer





















          • Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:26



















          1














          $$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$



          For $tan xne0,F(x)=dfrac{1-tan^2x}2$



          As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$






          share|cite|improve this answer





















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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            HINT



            We have that



            $$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$






            share|cite|improve this answer























            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27










            • Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
              – gimusi
              Nov 24 at 13:42
















            4














            HINT



            We have that



            $$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$






            share|cite|improve this answer























            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27










            • Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
              – gimusi
              Nov 24 at 13:42














            4












            4








            4






            HINT



            We have that



            $$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$






            share|cite|improve this answer














            HINT



            We have that



            $$cot (2x)cotleft(frac{pi }{2}-xright)=frac{cos(2x)}{sin(2x)}frac{sin x}{cos x}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 at 13:46

























            answered Nov 24 at 13:19









            gimusi

            1




            1












            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27










            • Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
              – gimusi
              Nov 24 at 13:42


















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27










            • Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
              – gimusi
              Nov 24 at 13:42
















            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:27




            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:27












            Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
            – gimusi
            Nov 24 at 13:42




            Indeed that’s the key point to recognize if you know the standard limit sinx/x. In general, it’s often good to reduce to sin x and cos x the expression.
            – gimusi
            Nov 24 at 13:42











            3














            Hint:
            $$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$






            share|cite|improve this answer





















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:28
















            3














            Hint:
            $$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$






            share|cite|improve this answer





















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:28














            3












            3








            3






            Hint:
            $$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$






            share|cite|improve this answer












            Hint:
            $$lim_{x to 0} cot (2x)cot (frac{pi }{2}-x)=lim_{x to 0} cot (2x)tan x=lim_{x to 0} dfrac{cos2x}{sin2x}dfrac{sin x}{cos x}=lim_{x to 0} dfrac{cos2x}{1}dfrac{2x}{sin2x}dfrac{sin x}{x}dfrac{1}{cos x}dfrac12$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 13:18









            Nosrati

            26.4k62353




            26.4k62353












            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:28


















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:28
















            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:28




            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:28











            3














            $displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$



            $displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $






            share|cite|improve this answer





















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27
















            3














            $displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$



            $displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $






            share|cite|improve this answer





















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27














            3












            3








            3






            $displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$



            $displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $






            share|cite|improve this answer












            $displaystyle cot 2xcot left(frac{pi }{2}-xright)=cot (2x)tan (x)=dfrac{cos 2xcdot sin x}{2sin xcos^2 x}=dfrac{cos 2x}{2cos^2x}$



            $displaystylelim_{xto0} cot 2xcot left(frac{pi }{2}-xright)=displaystylelim_{xto0}dfrac{cos 2x}{2cos^2x}=dfrac12 $







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 13:20









            Yadati Kiran

            1,702519




            1,702519












            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27


















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:27
















            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:27




            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:27











            2














            Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
            $$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$






            share|cite|improve this answer





















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:26
















            2














            Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
            $$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$






            share|cite|improve this answer





















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:26














            2












            2








            2






            Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
            $$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$






            share|cite|improve this answer












            Hint: $$lim_{x to 0} (cot (2x)cot (frac{pi }{2}-x))=lim_{x to 0} {cos (2x)over sin 2x}tan (x) =lim_{x to 0} {cos (2x)over 2sin xcos x}{sin xover cos x}$$
            $$=lim_{x to 0} {cos (2x)over 2cos^2 x} = {1over 2}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 13:17









            greedoid

            37.5k114794




            37.5k114794












            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:26


















            • Thanks, that tan(x) identity was what saved me :D
              – Jakcjones
              Nov 24 at 13:26
















            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:26




            Thanks, that tan(x) identity was what saved me :D
            – Jakcjones
            Nov 24 at 13:26











            1














            $$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$



            For $tan xne0,F(x)=dfrac{1-tan^2x}2$



            As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$






            share|cite|improve this answer


























              1














              $$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$



              For $tan xne0,F(x)=dfrac{1-tan^2x}2$



              As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$






              share|cite|improve this answer
























                1












                1








                1






                $$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$



                For $tan xne0,F(x)=dfrac{1-tan^2x}2$



                As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$






                share|cite|improve this answer












                $$F(x)=cot2xcotleft(dfracpi2-xright)=dfrac{tan x}{tan2x}=dfrac{tan x(1-tan^2x)}{2tan x}$$



                For $tan xne0,F(x)=dfrac{1-tan^2x}2$



                As $xto0,xne0implieslim_{xto0}F(x)=lim_{xto0}dfrac{1-tan^2x}2=?$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 6:49









                lab bhattacharjee

                222k15156274




                222k15156274






























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