Calculate $lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite
2












While working on limits in math my teacher gave us limits involving square roots as the following




Calculate:$$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n$$




and even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.



My own solution




$$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n~fracsqrt5n^2+4~+~sqrt5n^2+nsqrt5n^2+4~+~sqrt5n^2+n$$
$$lim_ntoinfty frac5n^2+4~-~(5n^2+n)sqrt5n^2+4~+~sqrt5n^2+n~=~lim_ntoinfty frac4-nnleft(sqrt5+frac4n^2~+~sqrt5+frac1nright)$$
$$=~frac-12sqrt5~=~-fracsqrt510$$





calculus limits roots




share|cite|improve this question

























    up vote
    2
    down vote

    favorite
    2












    While working on limits in math my teacher gave us limits involving square roots as the following




    Calculate:$$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n$$




    and even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.



    My own solution




    $$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n~fracsqrt5n^2+4~+~sqrt5n^2+nsqrt5n^2+4~+~sqrt5n^2+n$$
    $$lim_ntoinfty frac5n^2+4~-~(5n^2+n)sqrt5n^2+4~+~sqrt5n^2+n~=~lim_ntoinfty frac4-nnleft(sqrt5+frac4n^2~+~sqrt5+frac1nright)$$
    $$=~frac-12sqrt5~=~-fracsqrt510$$





    calculus limits roots




    share|cite|improve this question























      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      While working on limits in math my teacher gave us limits involving square roots as the following




      Calculate:$$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n$$




      and even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.



      My own solution




      $$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n~fracsqrt5n^2+4~+~sqrt5n^2+nsqrt5n^2+4~+~sqrt5n^2+n$$
      $$lim_ntoinfty frac5n^2+4~-~(5n^2+n)sqrt5n^2+4~+~sqrt5n^2+n~=~lim_ntoinfty frac4-nnleft(sqrt5+frac4n^2~+~sqrt5+frac1nright)$$
      $$=~frac-12sqrt5~=~-fracsqrt510$$





      calculus limits roots




      share|cite|improve this question













      While working on limits in math my teacher gave us limits involving square roots as the following




      Calculate:$$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n$$




      and even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.



      My own solution




      $$lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n~fracsqrt5n^2+4~+~sqrt5n^2+nsqrt5n^2+4~+~sqrt5n^2+n$$
      $$lim_ntoinfty frac5n^2+4~-~(5n^2+n)sqrt5n^2+4~+~sqrt5n^2+n~=~lim_ntoinfty frac4-nnleft(sqrt5+frac4n^2~+~sqrt5+frac1nright)$$
      $$=~frac-12sqrt5~=~-fracsqrt510$$






      calculus limits roots





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 23 at 9:14









      N. F. Taussig

      38.2k93053




      38.2k93053









      asked Jul 22 at 10:55









      mrtaurho

      740219




      740219




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          Two alternative ideas:



          (1) Write
          $$
          sqrt5n^2+4 - sqrt5n^2+n =
          sqrt 5n(sqrt1 + 4/5n^2 - sqrt1 + 1/5n)
          $$
          and use Taylor ($sqrt1 + x = 1 + x/2 - x^2/8 + cdots$).



          (2) Let be $f(x) = sqrt x$. By the Mean Value Theorem, for some $c_nin(5n^2 + 4,5n^2 + n)$
          $$
          sqrt5n^2 + 4 - sqrt5n^2 + n =
          f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
          -frac4 - n2sqrtc_n =
          -frac4/n - 12sqrtc_n/n^2,
          $$
          and by sqeezing $c_n/n^2to 5$.






          share|cite|improve this answer



















          • 3




            It is amazing to use mean value theorem to solve this problem.
            – Szeto
            Jul 22 at 15:33

















          up vote
          2
          down vote













          Let $1/n=h$



          As $ntoinfty,hto0+,h>0$



          $$sqrt5n^2+4=sqrtdfrac5+4h^2h^2=dfracsqrt5+4h^2sqrth^2$$



          Now $sqrth^2=|h|=+h$ for $h>0$



          So, we ahve
          $$lim_hto0^+dfracsqrt5+4h^2-sqrt5+hh=lim_hto0^+dfrac1sqrt5+4h^2+sqrt5+hcdotlim_hto0^+dfrac5+4h^2-(5+h)h=?$$






          share|cite|improve this answer

















          • 1




            This is more or less the same as OP’s solution.
            – Szeto
            Jul 22 at 11:20







          • 1




            @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
            – lab bhattacharjee
            Jul 22 at 11:22










          • I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
            – mrtaurho
            Jul 22 at 11:28

















          up vote
          1
          down vote













          Considering that $frac4n^2<< frac1n$ for big $n$ we have



          $$
          lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n equiv lim_hto 0fracsqrt5+4h^2-sqrt5+hh = lim_hto 0fracsqrt5+o(h^2)-sqrt5+hh
          $$



          now remember the derivative definition



          $$
          lim_hto 0fracsqrtx+h-sqrt xh = frac 12fracsqrt xx
          $$



          so the result is



          $$
          -frac 12fracsqrt 55
          $$






          share|cite|improve this answer





















            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );








             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859281%2fcalculate-lim-n-to-infty-sqrt5n24-sqrt5n2n%23new-answer', 'question_page');

            );

            Post as a guest

















            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Two alternative ideas:



            (1) Write
            $$
            sqrt5n^2+4 - sqrt5n^2+n =
            sqrt 5n(sqrt1 + 4/5n^2 - sqrt1 + 1/5n)
            $$
            and use Taylor ($sqrt1 + x = 1 + x/2 - x^2/8 + cdots$).



            (2) Let be $f(x) = sqrt x$. By the Mean Value Theorem, for some $c_nin(5n^2 + 4,5n^2 + n)$
            $$
            sqrt5n^2 + 4 - sqrt5n^2 + n =
            f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
            -frac4 - n2sqrtc_n =
            -frac4/n - 12sqrtc_n/n^2,
            $$
            and by sqeezing $c_n/n^2to 5$.






            share|cite|improve this answer



















            • 3




              It is amazing to use mean value theorem to solve this problem.
              – Szeto
              Jul 22 at 15:33














            up vote
            2
            down vote



            accepted










            Two alternative ideas:



            (1) Write
            $$
            sqrt5n^2+4 - sqrt5n^2+n =
            sqrt 5n(sqrt1 + 4/5n^2 - sqrt1 + 1/5n)
            $$
            and use Taylor ($sqrt1 + x = 1 + x/2 - x^2/8 + cdots$).



            (2) Let be $f(x) = sqrt x$. By the Mean Value Theorem, for some $c_nin(5n^2 + 4,5n^2 + n)$
            $$
            sqrt5n^2 + 4 - sqrt5n^2 + n =
            f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
            -frac4 - n2sqrtc_n =
            -frac4/n - 12sqrtc_n/n^2,
            $$
            and by sqeezing $c_n/n^2to 5$.






            share|cite|improve this answer



















            • 3




              It is amazing to use mean value theorem to solve this problem.
              – Szeto
              Jul 22 at 15:33












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Two alternative ideas:



            (1) Write
            $$
            sqrt5n^2+4 - sqrt5n^2+n =
            sqrt 5n(sqrt1 + 4/5n^2 - sqrt1 + 1/5n)
            $$
            and use Taylor ($sqrt1 + x = 1 + x/2 - x^2/8 + cdots$).



            (2) Let be $f(x) = sqrt x$. By the Mean Value Theorem, for some $c_nin(5n^2 + 4,5n^2 + n)$
            $$
            sqrt5n^2 + 4 - sqrt5n^2 + n =
            f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
            -frac4 - n2sqrtc_n =
            -frac4/n - 12sqrtc_n/n^2,
            $$
            and by sqeezing $c_n/n^2to 5$.






            share|cite|improve this answer















            Two alternative ideas:



            (1) Write
            $$
            sqrt5n^2+4 - sqrt5n^2+n =
            sqrt 5n(sqrt1 + 4/5n^2 - sqrt1 + 1/5n)
            $$
            and use Taylor ($sqrt1 + x = 1 + x/2 - x^2/8 + cdots$).



            (2) Let be $f(x) = sqrt x$. By the Mean Value Theorem, for some $c_nin(5n^2 + 4,5n^2 + n)$
            $$
            sqrt5n^2 + 4 - sqrt5n^2 + n =
            f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
            -frac4 - n2sqrtc_n =
            -frac4/n - 12sqrtc_n/n^2,
            $$
            and by sqeezing $c_n/n^2to 5$.







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 22 at 13:11


























            answered Jul 22 at 11:18









            Martín-Blas Pérez Pinilla

            33.4k42570




            33.4k42570







            • 3




              It is amazing to use mean value theorem to solve this problem.
              – Szeto
              Jul 22 at 15:33












            • 3




              It is amazing to use mean value theorem to solve this problem.
              – Szeto
              Jul 22 at 15:33







            3




            3




            It is amazing to use mean value theorem to solve this problem.
            – Szeto
            Jul 22 at 15:33




            It is amazing to use mean value theorem to solve this problem.
            – Szeto
            Jul 22 at 15:33










            up vote
            2
            down vote













            Let $1/n=h$



            As $ntoinfty,hto0+,h>0$



            $$sqrt5n^2+4=sqrtdfrac5+4h^2h^2=dfracsqrt5+4h^2sqrth^2$$



            Now $sqrth^2=|h|=+h$ for $h>0$



            So, we ahve
            $$lim_hto0^+dfracsqrt5+4h^2-sqrt5+hh=lim_hto0^+dfrac1sqrt5+4h^2+sqrt5+hcdotlim_hto0^+dfrac5+4h^2-(5+h)h=?$$






            share|cite|improve this answer

















            • 1




              This is more or less the same as OP’s solution.
              – Szeto
              Jul 22 at 11:20







            • 1




              @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
              – lab bhattacharjee
              Jul 22 at 11:22










            • I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
              – mrtaurho
              Jul 22 at 11:28














            up vote
            2
            down vote













            Let $1/n=h$



            As $ntoinfty,hto0+,h>0$



            $$sqrt5n^2+4=sqrtdfrac5+4h^2h^2=dfracsqrt5+4h^2sqrth^2$$



            Now $sqrth^2=|h|=+h$ for $h>0$



            So, we ahve
            $$lim_hto0^+dfracsqrt5+4h^2-sqrt5+hh=lim_hto0^+dfrac1sqrt5+4h^2+sqrt5+hcdotlim_hto0^+dfrac5+4h^2-(5+h)h=?$$






            share|cite|improve this answer

















            • 1




              This is more or less the same as OP’s solution.
              – Szeto
              Jul 22 at 11:20







            • 1




              @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
              – lab bhattacharjee
              Jul 22 at 11:22










            • I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
              – mrtaurho
              Jul 22 at 11:28












            up vote
            2
            down vote










            up vote
            2
            down vote









            Let $1/n=h$



            As $ntoinfty,hto0+,h>0$



            $$sqrt5n^2+4=sqrtdfrac5+4h^2h^2=dfracsqrt5+4h^2sqrth^2$$



            Now $sqrth^2=|h|=+h$ for $h>0$



            So, we ahve
            $$lim_hto0^+dfracsqrt5+4h^2-sqrt5+hh=lim_hto0^+dfrac1sqrt5+4h^2+sqrt5+hcdotlim_hto0^+dfrac5+4h^2-(5+h)h=?$$






            share|cite|improve this answer













            Let $1/n=h$



            As $ntoinfty,hto0+,h>0$



            $$sqrt5n^2+4=sqrtdfrac5+4h^2h^2=dfracsqrt5+4h^2sqrth^2$$



            Now $sqrth^2=|h|=+h$ for $h>0$



            So, we ahve
            $$lim_hto0^+dfracsqrt5+4h^2-sqrt5+hh=lim_hto0^+dfrac1sqrt5+4h^2+sqrt5+hcdotlim_hto0^+dfrac5+4h^2-(5+h)h=?$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 22 at 11:17









            lab bhattacharjee

            215k14152264




            215k14152264







            • 1




              This is more or less the same as OP’s solution.
              – Szeto
              Jul 22 at 11:20







            • 1




              @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
              – lab bhattacharjee
              Jul 22 at 11:22










            • I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
              – mrtaurho
              Jul 22 at 11:28












            • 1




              This is more or less the same as OP’s solution.
              – Szeto
              Jul 22 at 11:20







            • 1




              @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
              – lab bhattacharjee
              Jul 22 at 11:22










            • I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
              – mrtaurho
              Jul 22 at 11:28







            1




            1




            This is more or less the same as OP’s solution.
            – Szeto
            Jul 22 at 11:20





            This is more or less the same as OP’s solution.
            – Szeto
            Jul 22 at 11:20





            1




            1




            @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
            – lab bhattacharjee
            Jul 22 at 11:22




            @Szeto, Agreed. But I hope $lim hto0$ make things more explicit
            – lab bhattacharjee
            Jul 22 at 11:22












            I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
            – mrtaurho
            Jul 22 at 11:28




            I guess your approach @lab bhattacharjee is similiar to the two kinds of definition for $e$ : $e~=~lim_ntoinftyleft(1+frac1nright)^n~=~lim_hto 0left(1+hright)^frac1h$. You just reversed the way how to reach infinity or respectively zero. But yes, you are right it is more explicit to see in your solution but not exactly what I am looking for.
            – mrtaurho
            Jul 22 at 11:28










            up vote
            1
            down vote













            Considering that $frac4n^2<< frac1n$ for big $n$ we have



            $$
            lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n equiv lim_hto 0fracsqrt5+4h^2-sqrt5+hh = lim_hto 0fracsqrt5+o(h^2)-sqrt5+hh
            $$



            now remember the derivative definition



            $$
            lim_hto 0fracsqrtx+h-sqrt xh = frac 12fracsqrt xx
            $$



            so the result is



            $$
            -frac 12fracsqrt 55
            $$






            share|cite|improve this answer

























              up vote
              1
              down vote













              Considering that $frac4n^2<< frac1n$ for big $n$ we have



              $$
              lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n equiv lim_hto 0fracsqrt5+4h^2-sqrt5+hh = lim_hto 0fracsqrt5+o(h^2)-sqrt5+hh
              $$



              now remember the derivative definition



              $$
              lim_hto 0fracsqrtx+h-sqrt xh = frac 12fracsqrt xx
              $$



              so the result is



              $$
              -frac 12fracsqrt 55
              $$






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Considering that $frac4n^2<< frac1n$ for big $n$ we have



                $$
                lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n equiv lim_hto 0fracsqrt5+4h^2-sqrt5+hh = lim_hto 0fracsqrt5+o(h^2)-sqrt5+hh
                $$



                now remember the derivative definition



                $$
                lim_hto 0fracsqrtx+h-sqrt xh = frac 12fracsqrt xx
                $$



                so the result is



                $$
                -frac 12fracsqrt 55
                $$






                share|cite|improve this answer













                Considering that $frac4n^2<< frac1n$ for big $n$ we have



                $$
                lim_ntoinfty sqrt5n^2+4~-~sqrt5n^2+n equiv lim_hto 0fracsqrt5+4h^2-sqrt5+hh = lim_hto 0fracsqrt5+o(h^2)-sqrt5+hh
                $$



                now remember the derivative definition



                $$
                lim_hto 0fracsqrtx+h-sqrt xh = frac 12fracsqrt xx
                $$



                so the result is



                $$
                -frac 12fracsqrt 55
                $$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 22 at 11:30









                Cesareo

                5,7252412




                5,7252412






















                     

                    draft saved


                    draft discarded


























                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859281%2fcalculate-lim-n-to-infty-sqrt5n24-sqrt5n2n%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    What is the equation of a 3D cone with generalised tilt?

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                    Read more

                    Color the edges and diagonals of a regular polygon

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                    Read more

                    Relationship between determinant of matrix and determinant of adjoint?

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                    Read more