Limit Question: a better way to solve this limit? where $x to -infty$












3












$begingroup$


Hey guys so I have this limit:



$$lim_{x to -∞} f(x) = {(x+sqrt{x^2+2x})}$$



I solved it by multiplying numerator and denominator by $$x-sqrt{x^2+2x}$$ and got $-1$ as my answer, but I really don't like how I solved it; any better way to solve it?






limits







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$endgroup$

















    3












    $begingroup$


    Hey guys so I have this limit:



    $$lim_{x to -∞} f(x) = {(x+sqrt{x^2+2x})}$$



    I solved it by multiplying numerator and denominator by $$x-sqrt{x^2+2x}$$ and got $-1$ as my answer, but I really don't like how I solved it; any better way to solve it?






    limits







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Hey guys so I have this limit:



      $$lim_{x to -∞} f(x) = {(x+sqrt{x^2+2x})}$$



      I solved it by multiplying numerator and denominator by $$x-sqrt{x^2+2x}$$ and got $-1$ as my answer, but I really don't like how I solved it; any better way to solve it?






      limits







      share|cite|improve this question









      $endgroup$




      Hey guys so I have this limit:



      $$lim_{x to -∞} f(x) = {(x+sqrt{x^2+2x})}$$



      I solved it by multiplying numerator and denominator by $$x-sqrt{x^2+2x}$$ and got $-1$ as my answer, but I really don't like how I solved it; any better way to solve it?






      limits




      limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 5 '18 at 22:10









      S..S..

      545




      545






















          4 Answers
          4






          active

          oldest

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          3












          $begingroup$

          You can set $x=-dfrac1tenspace(t>0,,tto 0)$. The expression rewrites as
          $$f(x)=-frac1t+sqrt{frac{1mathstrut{}}{t^2}-frac2t}=frac{-1+sqrt{strut 1-2t}}t.$$
          This is the rate of variation of the function $sqrt{1-2t};$ from $t=0$, so the limit is
          $$bigl(sqrt{1-2t}bigr)'_{t=0}=frac1{2sqrt{1-2t}}cdot(-2)Biggm|_{t=0}=-1.$$






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            You are looking, setting $t=|x|$, at
            $$
            lim_{ttoinfty} sqrt{t^2-2t} - t
            $$
            For the sake of it, here is a solution using the Taylor expansion of $sqrt{1+u}$ around $0$: for $t>0$,
            $$
            sqrt{t^2-2t} - t
            = tsqrt{1-frac{2}{t}} - t
            = tleft(sqrt{1-frac{2}{t}} - 1right)
            = tleft(1-frac{1}{t} + oleft(frac{1}{t}right) - 1right)
            = -1+o(1)
            $$
            showing the limit is indeed $-1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I don't know what you expect me to say...
              $endgroup$
              – Clement C.
              Dec 6 '18 at 20:21



















            0












            $begingroup$

            For fun



            $t=-x$, let $t >2.$



            $(t^2-2t+1-1)^{1/2}-t=$



            $((t-1)^2-1)^{1/2} -((t-1)^2)^{1/2}-1;$



            Then with $m:=(t-1)^2$.



            $(m-1)^{1/2}- m^{1/2} -1$.



            $y_m:=(m-1)^{1/2}-m^{1/2}.$



            $lim_{m rightarrow infty} y_m=0$.



            Can you prove it?






            share|cite|improve this answer









            $endgroup$





















              -2












              $begingroup$

              I suggest to flip the sign by $y=-xto infty$ then



              $$lim_{x to -∞} f(x) = {x+sqrt{x^2+2x}}=lim_{y to infty} f(x) = {-y+sqrt{y^2-2y}}$$



              and then



              $$sqrt{y^2-2y}-y=left(sqrt{y^2-2y}-yright)cdotfrac{sqrt{y^2-2y}+y}{sqrt{y^2-2y}+y}$$



              Edit As another alternative we can use binomial series $tto 0, ,(1+t)^n=1+nt+O(t^2)$, are you aware about that?



              We proceed as follows



              $$sqrt{y^2-2y}=sqrt{y^2}cdot sqrt{1-2/y}=ycdot(1-2/y)^frac12=ycdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$



              then



              $$sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) to -1$$



              Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then



              $$|O(1/y)|le frac c y to 0 implies O(1/y)to 0$$






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Not aware of that one, could you explain?
                $endgroup$
                – S..
                Dec 5 '18 at 22:16










              • $begingroup$
                @S.. Ok I'll add some details for the second limit.
                $endgroup$
                – gimusi
                Dec 5 '18 at 22:17










              • $begingroup$
                @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                $endgroup$
                – gimusi
                Dec 5 '18 at 22:21










              • $begingroup$
                @amWhy Thanks for the editing! it's nice now
                $endgroup$
                – gimusi
                Dec 5 '18 at 22:22










              • $begingroup$
                @gimusi yeah that term is new for me; would be great if you help me out with that
                $endgroup$
                – S..
                Dec 5 '18 at 22:22











              Your Answer





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              4 Answers
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              4 Answers
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              3












              $begingroup$

              You can set $x=-dfrac1tenspace(t>0,,tto 0)$. The expression rewrites as
              $$f(x)=-frac1t+sqrt{frac{1mathstrut{}}{t^2}-frac2t}=frac{-1+sqrt{strut 1-2t}}t.$$
              This is the rate of variation of the function $sqrt{1-2t};$ from $t=0$, so the limit is
              $$bigl(sqrt{1-2t}bigr)'_{t=0}=frac1{2sqrt{1-2t}}cdot(-2)Biggm|_{t=0}=-1.$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                You can set $x=-dfrac1tenspace(t>0,,tto 0)$. The expression rewrites as
                $$f(x)=-frac1t+sqrt{frac{1mathstrut{}}{t^2}-frac2t}=frac{-1+sqrt{strut 1-2t}}t.$$
                This is the rate of variation of the function $sqrt{1-2t};$ from $t=0$, so the limit is
                $$bigl(sqrt{1-2t}bigr)'_{t=0}=frac1{2sqrt{1-2t}}cdot(-2)Biggm|_{t=0}=-1.$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You can set $x=-dfrac1tenspace(t>0,,tto 0)$. The expression rewrites as
                  $$f(x)=-frac1t+sqrt{frac{1mathstrut{}}{t^2}-frac2t}=frac{-1+sqrt{strut 1-2t}}t.$$
                  This is the rate of variation of the function $sqrt{1-2t};$ from $t=0$, so the limit is
                  $$bigl(sqrt{1-2t}bigr)'_{t=0}=frac1{2sqrt{1-2t}}cdot(-2)Biggm|_{t=0}=-1.$$






                  share|cite|improve this answer









                  $endgroup$



                  You can set $x=-dfrac1tenspace(t>0,,tto 0)$. The expression rewrites as
                  $$f(x)=-frac1t+sqrt{frac{1mathstrut{}}{t^2}-frac2t}=frac{-1+sqrt{strut 1-2t}}t.$$
                  This is the rate of variation of the function $sqrt{1-2t};$ from $t=0$, so the limit is
                  $$bigl(sqrt{1-2t}bigr)'_{t=0}=frac1{2sqrt{1-2t}}cdot(-2)Biggm|_{t=0}=-1.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 22:29









                  BernardBernard

                  119k740113




                  119k740113























                      3












                      $begingroup$

                      You are looking, setting $t=|x|$, at
                      $$
                      lim_{ttoinfty} sqrt{t^2-2t} - t
                      $$
                      For the sake of it, here is a solution using the Taylor expansion of $sqrt{1+u}$ around $0$: for $t>0$,
                      $$
                      sqrt{t^2-2t} - t
                      = tsqrt{1-frac{2}{t}} - t
                      = tleft(sqrt{1-frac{2}{t}} - 1right)
                      = tleft(1-frac{1}{t} + oleft(frac{1}{t}right) - 1right)
                      = -1+o(1)
                      $$
                      showing the limit is indeed $-1$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I don't know what you expect me to say...
                        $endgroup$
                        – Clement C.
                        Dec 6 '18 at 20:21
















                      3












                      $begingroup$

                      You are looking, setting $t=|x|$, at
                      $$
                      lim_{ttoinfty} sqrt{t^2-2t} - t
                      $$
                      For the sake of it, here is a solution using the Taylor expansion of $sqrt{1+u}$ around $0$: for $t>0$,
                      $$
                      sqrt{t^2-2t} - t
                      = tsqrt{1-frac{2}{t}} - t
                      = tleft(sqrt{1-frac{2}{t}} - 1right)
                      = tleft(1-frac{1}{t} + oleft(frac{1}{t}right) - 1right)
                      = -1+o(1)
                      $$
                      showing the limit is indeed $-1$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        I don't know what you expect me to say...
                        $endgroup$
                        – Clement C.
                        Dec 6 '18 at 20:21














                      3












                      3








                      3





                      $begingroup$

                      You are looking, setting $t=|x|$, at
                      $$
                      lim_{ttoinfty} sqrt{t^2-2t} - t
                      $$
                      For the sake of it, here is a solution using the Taylor expansion of $sqrt{1+u}$ around $0$: for $t>0$,
                      $$
                      sqrt{t^2-2t} - t
                      = tsqrt{1-frac{2}{t}} - t
                      = tleft(sqrt{1-frac{2}{t}} - 1right)
                      = tleft(1-frac{1}{t} + oleft(frac{1}{t}right) - 1right)
                      = -1+o(1)
                      $$
                      showing the limit is indeed $-1$.






                      share|cite|improve this answer











                      $endgroup$



                      You are looking, setting $t=|x|$, at
                      $$
                      lim_{ttoinfty} sqrt{t^2-2t} - t
                      $$
                      For the sake of it, here is a solution using the Taylor expansion of $sqrt{1+u}$ around $0$: for $t>0$,
                      $$
                      sqrt{t^2-2t} - t
                      = tsqrt{1-frac{2}{t}} - t
                      = tleft(sqrt{1-frac{2}{t}} - 1right)
                      = tleft(1-frac{1}{t} + oleft(frac{1}{t}right) - 1right)
                      = -1+o(1)
                      $$
                      showing the limit is indeed $-1$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 5 '18 at 22:23

























                      answered Dec 5 '18 at 22:20









                      Clement C.Clement C.

                      49.8k33887




                      49.8k33887












                      • $begingroup$
                        I don't know what you expect me to say...
                        $endgroup$
                        – Clement C.
                        Dec 6 '18 at 20:21


















                      • $begingroup$
                        I don't know what you expect me to say...
                        $endgroup$
                        – Clement C.
                        Dec 6 '18 at 20:21
















                      $begingroup$
                      I don't know what you expect me to say...
                      $endgroup$
                      – Clement C.
                      Dec 6 '18 at 20:21




                      $begingroup$
                      I don't know what you expect me to say...
                      $endgroup$
                      – Clement C.
                      Dec 6 '18 at 20:21











                      0












                      $begingroup$

                      For fun



                      $t=-x$, let $t >2.$



                      $(t^2-2t+1-1)^{1/2}-t=$



                      $((t-1)^2-1)^{1/2} -((t-1)^2)^{1/2}-1;$



                      Then with $m:=(t-1)^2$.



                      $(m-1)^{1/2}- m^{1/2} -1$.



                      $y_m:=(m-1)^{1/2}-m^{1/2}.$



                      $lim_{m rightarrow infty} y_m=0$.



                      Can you prove it?






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        For fun



                        $t=-x$, let $t >2.$



                        $(t^2-2t+1-1)^{1/2}-t=$



                        $((t-1)^2-1)^{1/2} -((t-1)^2)^{1/2}-1;$



                        Then with $m:=(t-1)^2$.



                        $(m-1)^{1/2}- m^{1/2} -1$.



                        $y_m:=(m-1)^{1/2}-m^{1/2}.$



                        $lim_{m rightarrow infty} y_m=0$.



                        Can you prove it?






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          For fun



                          $t=-x$, let $t >2.$



                          $(t^2-2t+1-1)^{1/2}-t=$



                          $((t-1)^2-1)^{1/2} -((t-1)^2)^{1/2}-1;$



                          Then with $m:=(t-1)^2$.



                          $(m-1)^{1/2}- m^{1/2} -1$.



                          $y_m:=(m-1)^{1/2}-m^{1/2}.$



                          $lim_{m rightarrow infty} y_m=0$.



                          Can you prove it?






                          share|cite|improve this answer









                          $endgroup$



                          For fun



                          $t=-x$, let $t >2.$



                          $(t^2-2t+1-1)^{1/2}-t=$



                          $((t-1)^2-1)^{1/2} -((t-1)^2)^{1/2}-1;$



                          Then with $m:=(t-1)^2$.



                          $(m-1)^{1/2}- m^{1/2} -1$.



                          $y_m:=(m-1)^{1/2}-m^{1/2}.$



                          $lim_{m rightarrow infty} y_m=0$.



                          Can you prove it?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 5 '18 at 22:58









                          Peter SzilasPeter Szilas

                          11.1k2821




                          11.1k2821























                              -2












                              $begingroup$

                              I suggest to flip the sign by $y=-xto infty$ then



                              $$lim_{x to -∞} f(x) = {x+sqrt{x^2+2x}}=lim_{y to infty} f(x) = {-y+sqrt{y^2-2y}}$$



                              and then



                              $$sqrt{y^2-2y}-y=left(sqrt{y^2-2y}-yright)cdotfrac{sqrt{y^2-2y}+y}{sqrt{y^2-2y}+y}$$



                              Edit As another alternative we can use binomial series $tto 0, ,(1+t)^n=1+nt+O(t^2)$, are you aware about that?



                              We proceed as follows



                              $$sqrt{y^2-2y}=sqrt{y^2}cdot sqrt{1-2/y}=ycdot(1-2/y)^frac12=ycdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$



                              then



                              $$sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) to -1$$



                              Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then



                              $$|O(1/y)|le frac c y to 0 implies O(1/y)to 0$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                Not aware of that one, could you explain?
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:16










                              • $begingroup$
                                @S.. Ok I'll add some details for the second limit.
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:17










                              • $begingroup$
                                @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:21










                              • $begingroup$
                                @amWhy Thanks for the editing! it's nice now
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:22










                              • $begingroup$
                                @gimusi yeah that term is new for me; would be great if you help me out with that
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:22
















                              -2












                              $begingroup$

                              I suggest to flip the sign by $y=-xto infty$ then



                              $$lim_{x to -∞} f(x) = {x+sqrt{x^2+2x}}=lim_{y to infty} f(x) = {-y+sqrt{y^2-2y}}$$



                              and then



                              $$sqrt{y^2-2y}-y=left(sqrt{y^2-2y}-yright)cdotfrac{sqrt{y^2-2y}+y}{sqrt{y^2-2y}+y}$$



                              Edit As another alternative we can use binomial series $tto 0, ,(1+t)^n=1+nt+O(t^2)$, are you aware about that?



                              We proceed as follows



                              $$sqrt{y^2-2y}=sqrt{y^2}cdot sqrt{1-2/y}=ycdot(1-2/y)^frac12=ycdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$



                              then



                              $$sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) to -1$$



                              Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then



                              $$|O(1/y)|le frac c y to 0 implies O(1/y)to 0$$






                              share|cite|improve this answer











                              $endgroup$













                              • $begingroup$
                                Not aware of that one, could you explain?
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:16










                              • $begingroup$
                                @S.. Ok I'll add some details for the second limit.
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:17










                              • $begingroup$
                                @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:21










                              • $begingroup$
                                @amWhy Thanks for the editing! it's nice now
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:22










                              • $begingroup$
                                @gimusi yeah that term is new for me; would be great if you help me out with that
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:22














                              -2












                              -2








                              -2





                              $begingroup$

                              I suggest to flip the sign by $y=-xto infty$ then



                              $$lim_{x to -∞} f(x) = {x+sqrt{x^2+2x}}=lim_{y to infty} f(x) = {-y+sqrt{y^2-2y}}$$



                              and then



                              $$sqrt{y^2-2y}-y=left(sqrt{y^2-2y}-yright)cdotfrac{sqrt{y^2-2y}+y}{sqrt{y^2-2y}+y}$$



                              Edit As another alternative we can use binomial series $tto 0, ,(1+t)^n=1+nt+O(t^2)$, are you aware about that?



                              We proceed as follows



                              $$sqrt{y^2-2y}=sqrt{y^2}cdot sqrt{1-2/y}=ycdot(1-2/y)^frac12=ycdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$



                              then



                              $$sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) to -1$$



                              Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then



                              $$|O(1/y)|le frac c y to 0 implies O(1/y)to 0$$






                              share|cite|improve this answer











                              $endgroup$



                              I suggest to flip the sign by $y=-xto infty$ then



                              $$lim_{x to -∞} f(x) = {x+sqrt{x^2+2x}}=lim_{y to infty} f(x) = {-y+sqrt{y^2-2y}}$$



                              and then



                              $$sqrt{y^2-2y}-y=left(sqrt{y^2-2y}-yright)cdotfrac{sqrt{y^2-2y}+y}{sqrt{y^2-2y}+y}$$



                              Edit As another alternative we can use binomial series $tto 0, ,(1+t)^n=1+nt+O(t^2)$, are you aware about that?



                              We proceed as follows



                              $$sqrt{y^2-2y}=sqrt{y^2}cdot sqrt{1-2/y}=ycdot(1-2/y)^frac12=ycdot(1-1/y+O(1/y^2))=y-1+O(1/y),$$



                              then



                              $$sqrt{y^2-2y}-y=y-1+O(1/y)-y=-1+O(1/y) to -1$$



                              Indeed the $O(1/y)$ notation collects all the terms in absolute value eventually smaller or equal than $c/y$ with $c$ some positive constant then



                              $$|O(1/y)|le frac c y to 0 implies O(1/y)to 0$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 5 '18 at 22:51

























                              answered Dec 5 '18 at 22:13









                              gimusigimusi

                              92.8k84494




                              92.8k84494












                              • $begingroup$
                                Not aware of that one, could you explain?
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:16










                              • $begingroup$
                                @S.. Ok I'll add some details for the second limit.
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:17










                              • $begingroup$
                                @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:21










                              • $begingroup$
                                @amWhy Thanks for the editing! it's nice now
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:22










                              • $begingroup$
                                @gimusi yeah that term is new for me; would be great if you help me out with that
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:22


















                              • $begingroup$
                                Not aware of that one, could you explain?
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:16










                              • $begingroup$
                                @S.. Ok I'll add some details for the second limit.
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:17










                              • $begingroup$
                                @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:21










                              • $begingroup$
                                @amWhy Thanks for the editing! it's nice now
                                $endgroup$
                                – gimusi
                                Dec 5 '18 at 22:22










                              • $begingroup$
                                @gimusi yeah that term is new for me; would be great if you help me out with that
                                $endgroup$
                                – S..
                                Dec 5 '18 at 22:22
















                              $begingroup$
                              Not aware of that one, could you explain?
                              $endgroup$
                              – S..
                              Dec 5 '18 at 22:16




                              $begingroup$
                              Not aware of that one, could you explain?
                              $endgroup$
                              – S..
                              Dec 5 '18 at 22:16












                              $begingroup$
                              @S.. Ok I'll add some details for the second limit.
                              $endgroup$
                              – gimusi
                              Dec 5 '18 at 22:17




                              $begingroup$
                              @S.. Ok I'll add some details for the second limit.
                              $endgroup$
                              – gimusi
                              Dec 5 '18 at 22:17












                              $begingroup$
                              @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                              $endgroup$
                              – gimusi
                              Dec 5 '18 at 22:21




                              $begingroup$
                              @S.. Do not hesitate to ask for any clarification on that! Maybe the little-o term is not clear to you? Are you more confortable with big-O notation? Let me know about that!
                              $endgroup$
                              – gimusi
                              Dec 5 '18 at 22:21












                              $begingroup$
                              @amWhy Thanks for the editing! it's nice now
                              $endgroup$
                              – gimusi
                              Dec 5 '18 at 22:22




                              $begingroup$
                              @amWhy Thanks for the editing! it's nice now
                              $endgroup$
                              – gimusi
                              Dec 5 '18 at 22:22












                              $begingroup$
                              @gimusi yeah that term is new for me; would be great if you help me out with that
                              $endgroup$
                              – S..
                              Dec 5 '18 at 22:22




                              $begingroup$
                              @gimusi yeah that term is new for me; would be great if you help me out with that
                              $endgroup$
                              – S..
                              Dec 5 '18 at 22:22


















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