Quickly evaluating this limitA limit problem related to $log sec x$Help in evaluating limit.Evaluate $Im left (frac1100times 2^100(e^2iota x -1)^100right )$Help evaluating this limitHow to quickly solve $y=int_-pi/4 ^pi/4 left[cos x + sqrt1+x^2sin^3xcos^3xright]dx$?Where am I going wrong in evaluating this limit?Evaluating complicated limit

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Quickly evaluating this limit


A limit problem related to $log sec x$Help in evaluating limit.Evaluate $Im left (frac1100times 2^100(e^2iota x -1)^100right )$Help evaluating this limitHow to quickly solve $y=int_-pi/4 ^pi/4 left[cos x + sqrt1+x^2sin^3xcos^3xright]dx$?Where am I going wrong in evaluating this limit?Evaluating complicated limit






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.



$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$



If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as



$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$



where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.










share|cite|improve this question











$endgroup$









  • 3




    $begingroup$
    As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
    $endgroup$
    – ganeshie8
    8 hours ago


















1












$begingroup$


I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.



$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$



If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as



$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$



where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.










share|cite|improve this question











$endgroup$









  • 3




    $begingroup$
    As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
    $endgroup$
    – ganeshie8
    8 hours ago














1












1








1


2



$begingroup$


I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.



$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$



If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as



$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$



where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.










share|cite|improve this question











$endgroup$




I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.



$$lim_xto 0left(frac1x^2-frac1sin^2 xright).$$



If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as



$$frac1x^2-frac1sin^2 x = fracsin^2 x - x^2x^2sin^2 x = left(fracx^2sin^2 xright)left(fracsin x + xxright)left(fracsin x - xx^3right),$$



where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.







calculus limits intuition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









José Carlos Santos

209k26 gold badges166 silver badges289 bronze badges




209k26 gold badges166 silver badges289 bronze badges










asked 9 hours ago









JosephSloteJosephSlote

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  • 3




    $begingroup$
    As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
    $endgroup$
    – ganeshie8
    8 hours ago













  • 3




    $begingroup$
    As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
    $endgroup$
    – ganeshie8
    8 hours ago








3




3




$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago





$begingroup$
As soon as you see $x^2-y^2$ don't you automatically do $(x+y)(x-y)$ ?
$endgroup$
– ganeshie8
8 hours ago











3 Answers
3






active

oldest

votes


















5











$begingroup$

Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.






share|cite|improve this answer









$endgroup$






















    4











    $begingroup$

    Once you know that
    $dfracsin xx
    to 1$
    ,
    and similar,
    fairly simple limits,
    it becomes natural
    to try to make them appear
    by extracting them
    from more complicated expressions.






    share|cite|improve this answer









    $endgroup$






















      3











      $begingroup$

      OK, let's talk through the thought process.



      • When I see a difference of two fractions, I give them common
        denominators first, as per your second $=$.


      • I can't help but factorise the new numerator's difference of two
        squares after that.


      • Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
        need to take out a $left(fracsin xxright)^-2$ factor before
        I can go any further.


      • That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
        number of ways I could write that as a product of two other factors,
        but I don't want to factor out a limit outside $Bbb
        Rsetminus0$
        , in case I get an indeterminate form. (That same
        concern motivation the previous handling of $sin^-2x$.) Well,
        $fracsin xxsim1impliesfracsin x+xxsim2$, so finally
        the sought limit is twice $lim_xto0fracsin x-xx^3$.






      share|cite|improve this answer









      $endgroup$










      • 1




        $begingroup$
        Nice detailed expression of what I just gave a cursory overview of.
        $endgroup$
        – marty cohen
        8 hours ago













      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5











      $begingroup$

      Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.






      share|cite|improve this answer









      $endgroup$



















        5











        $begingroup$

        Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.






        share|cite|improve this answer









        $endgroup$

















          5












          5








          5





          $begingroup$

          Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.






          share|cite|improve this answer









          $endgroup$



          Since$$sin(x)=x-fracx^33!+fracx^55!-cdots,$$you know that$$sin(x)+x=2x-fracx^33!+fracx^55!-cdots$$and that$$sin(x)-x=-fracx^33!+fracx^55!-cdots.$$Therefore both limits$$lim_xto0fracsin(x)+xxtext and lim_xto0fracsin(x)-xx^3$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          José Carlos SantosJosé Carlos Santos

          209k26 gold badges166 silver badges289 bronze badges




          209k26 gold badges166 silver badges289 bronze badges


























              4











              $begingroup$

              Once you know that
              $dfracsin xx
              to 1$
              ,
              and similar,
              fairly simple limits,
              it becomes natural
              to try to make them appear
              by extracting them
              from more complicated expressions.






              share|cite|improve this answer









              $endgroup$



















                4











                $begingroup$

                Once you know that
                $dfracsin xx
                to 1$
                ,
                and similar,
                fairly simple limits,
                it becomes natural
                to try to make them appear
                by extracting them
                from more complicated expressions.






                share|cite|improve this answer









                $endgroup$

















                  4












                  4








                  4





                  $begingroup$

                  Once you know that
                  $dfracsin xx
                  to 1$
                  ,
                  and similar,
                  fairly simple limits,
                  it becomes natural
                  to try to make them appear
                  by extracting them
                  from more complicated expressions.






                  share|cite|improve this answer









                  $endgroup$



                  Once you know that
                  $dfracsin xx
                  to 1$
                  ,
                  and similar,
                  fairly simple limits,
                  it becomes natural
                  to try to make them appear
                  by extracting them
                  from more complicated expressions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  marty cohenmarty cohen

                  79.4k5 gold badges52 silver badges134 bronze badges




                  79.4k5 gold badges52 silver badges134 bronze badges
























                      3











                      $begingroup$

                      OK, let's talk through the thought process.



                      • When I see a difference of two fractions, I give them common
                        denominators first, as per your second $=$.


                      • I can't help but factorise the new numerator's difference of two
                        squares after that.


                      • Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
                        need to take out a $left(fracsin xxright)^-2$ factor before
                        I can go any further.


                      • That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
                        number of ways I could write that as a product of two other factors,
                        but I don't want to factor out a limit outside $Bbb
                        Rsetminus0$
                        , in case I get an indeterminate form. (That same
                        concern motivation the previous handling of $sin^-2x$.) Well,
                        $fracsin xxsim1impliesfracsin x+xxsim2$, so finally
                        the sought limit is twice $lim_xto0fracsin x-xx^3$.






                      share|cite|improve this answer









                      $endgroup$










                      • 1




                        $begingroup$
                        Nice detailed expression of what I just gave a cursory overview of.
                        $endgroup$
                        – marty cohen
                        8 hours ago















                      3











                      $begingroup$

                      OK, let's talk through the thought process.



                      • When I see a difference of two fractions, I give them common
                        denominators first, as per your second $=$.


                      • I can't help but factorise the new numerator's difference of two
                        squares after that.


                      • Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
                        need to take out a $left(fracsin xxright)^-2$ factor before
                        I can go any further.


                      • That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
                        number of ways I could write that as a product of two other factors,
                        but I don't want to factor out a limit outside $Bbb
                        Rsetminus0$
                        , in case I get an indeterminate form. (That same
                        concern motivation the previous handling of $sin^-2x$.) Well,
                        $fracsin xxsim1impliesfracsin x+xxsim2$, so finally
                        the sought limit is twice $lim_xto0fracsin x-xx^3$.






                      share|cite|improve this answer









                      $endgroup$










                      • 1




                        $begingroup$
                        Nice detailed expression of what I just gave a cursory overview of.
                        $endgroup$
                        – marty cohen
                        8 hours ago













                      3












                      3








                      3





                      $begingroup$

                      OK, let's talk through the thought process.



                      • When I see a difference of two fractions, I give them common
                        denominators first, as per your second $=$.


                      • I can't help but factorise the new numerator's difference of two
                        squares after that.


                      • Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
                        need to take out a $left(fracsin xxright)^-2$ factor before
                        I can go any further.


                      • That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
                        number of ways I could write that as a product of two other factors,
                        but I don't want to factor out a limit outside $Bbb
                        Rsetminus0$
                        , in case I get an indeterminate form. (That same
                        concern motivation the previous handling of $sin^-2x$.) Well,
                        $fracsin xxsim1impliesfracsin x+xxsim2$, so finally
                        the sought limit is twice $lim_xto0fracsin x-xx^3$.






                      share|cite|improve this answer









                      $endgroup$



                      OK, let's talk through the thought process.



                      • When I see a difference of two fractions, I give them common
                        denominators first, as per your second $=$.


                      • I can't help but factorise the new numerator's difference of two
                        squares after that.


                      • Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
                        need to take out a $left(fracsin xxright)^-2$ factor before
                        I can go any further.


                      • That leaves me with $frac(sin x+x)(sin x-x)x^4$. There are any
                        number of ways I could write that as a product of two other factors,
                        but I don't want to factor out a limit outside $Bbb
                        Rsetminus0$
                        , in case I get an indeterminate form. (That same
                        concern motivation the previous handling of $sin^-2x$.) Well,
                        $fracsin xxsim1impliesfracsin x+xxsim2$, so finally
                        the sought limit is twice $lim_xto0fracsin x-xx^3$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 8 hours ago









                      J.G.J.G.

                      45k2 gold badges41 silver badges62 bronze badges




                      45k2 gold badges41 silver badges62 bronze badges










                      • 1




                        $begingroup$
                        Nice detailed expression of what I just gave a cursory overview of.
                        $endgroup$
                        – marty cohen
                        8 hours ago












                      • 1




                        $begingroup$
                        Nice detailed expression of what I just gave a cursory overview of.
                        $endgroup$
                        – marty cohen
                        8 hours ago







                      1




                      1




                      $begingroup$
                      Nice detailed expression of what I just gave a cursory overview of.
                      $endgroup$
                      – marty cohen
                      8 hours ago




                      $begingroup$
                      Nice detailed expression of what I just gave a cursory overview of.
                      $endgroup$
                      – marty cohen
                      8 hours ago

















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