Differentiability of $g(x)sin(1/x)$ at $x=0$.











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I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:




Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
$$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
is differentiable at $x=0$.




My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
begin{align*}
lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
\ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
end{align*}
Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
$$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
Hence, $f'(0)=0$.



I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.






real-analysis proof-verification







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    up vote
    0
    down vote

    favorite












    I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:




    Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
    $$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
    is differentiable at $x=0$.




    My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
    begin{align*}
    lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
    \ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
    end{align*}
    Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
    $$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
    Hence, $f'(0)=0$.



    I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.






    real-analysis proof-verification







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:




      Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
      $$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
      is differentiable at $x=0$.




      My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
      begin{align*}
      lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
      \ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
      end{align*}
      Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
      $$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
      Hence, $f'(0)=0$.



      I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.






      real-analysis proof-verification







      share|cite|improve this question













      I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:




      Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that
      $$f(x)=begin{cases} g(x)sin(x) && xneq 0 \ 0 && x=0end{cases}$$
      is differentiable at $x=0$.




      My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $lim_{hto 0}frac{f(h)-f(0)}{h}$ exists. We have that
      begin{align*}
      lim_{hto 0}frac{f(h)-f(0)}{h} & = lim_{hto 0} frac{g(h)sin(1/h)-0}{h}
      \ & = lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)
      end{align*}
      Since $g'(0)=0$, we have that $lim_{hto 0}frac{g(h)-g(0)}{h}=lim_{hto 0}frac{g(h)}{h}=0$. Therefore,
      $$lim_{hto 0}frac{g(h)}{h}cdotsin(1/h)=0cdotlim_{hto 0}sin(1/h)=0$$
      Hence, $f'(0)=0$.



      I am unsure whether I can say that $0cdotlim_{hto 0}sin(1/h)=0$.






      real-analysis proof-verification




      real-analysis proof-verification






      share|cite|improve this question













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      asked 23 hours ago









      高田航

      1,274318




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          4 Answers
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          up vote
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          down vote



          accepted










          The limit $lim_{hto 0}sin(1/h)$ does not exist !



          But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.






          share|cite|improve this answer




























            up vote
            2
            down vote













            Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.



            It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$






            share|cite|improve this answer




























              up vote
              1
              down vote













              If I am correct then the question mention's about



              $$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$



              So I would use the following:



              $$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$



              $$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$



              Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.



              This shortened the expression to



              $$ lim_{h rightarrow 0} g(h) $$



              Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.






              share|cite|improve this answer




























                up vote
                0
                down vote













                By Taylor's theorem:
                $$g(x)=g(0)+g'(0)x+r(x)x$$
                $$g(x)=xr(x)$$
                So
                $$f(x)=xr(x)sinleft(frac{1}{x}right)$$
                And
                $$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
                Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.






                share|cite|improve this answer





















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                  4 Answers
                  4






                  active

                  oldest

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                  4 Answers
                  4






                  active

                  oldest

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                  active

                  oldest

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                  active

                  oldest

                  votes








                  up vote
                  4
                  down vote



                  accepted










                  The limit $lim_{hto 0}sin(1/h)$ does not exist !



                  But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.






                  share|cite|improve this answer

























                    up vote
                    4
                    down vote



                    accepted










                    The limit $lim_{hto 0}sin(1/h)$ does not exist !



                    But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.






                    share|cite|improve this answer























                      up vote
                      4
                      down vote



                      accepted







                      up vote
                      4
                      down vote



                      accepted






                      The limit $lim_{hto 0}sin(1/h)$ does not exist !



                      But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.






                      share|cite|improve this answer












                      The limit $lim_{hto 0}sin(1/h)$ does not exist !



                      But since $|frac{g(h)}{h}cdotsin(1/h)| le |frac{g(h)}{h}|$ and $lim_{hto 0}frac{g(h)}{h}=0$, we see that $lim_{hto 0}frac{f(h)-f(0)}{h}=0$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 23 hours ago









                      Fred

                      41.9k1642




                      41.9k1642






















                          up vote
                          2
                          down vote













                          Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.



                          It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$






                          share|cite|improve this answer

























                            up vote
                            2
                            down vote













                            Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.



                            It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$






                            share|cite|improve this answer























                              up vote
                              2
                              down vote










                              up vote
                              2
                              down vote









                              Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.



                              It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$






                              share|cite|improve this answer












                              Yes, you can say that $0timeslim_{hto0}sinleft(frac1hright)=0$, but that's not a sensible thing to say.



                              It is better to say that$$leftlvertfrac{g(h)}hsinleft(frac1hright)rightrvertleqslantleftlvertfrac{g(h)}hrightrvert$$and then to uset the fact that$$lim_{hto0}frac{g(h)}h=0ifflim_{hto0}leftlvertfrac{g(h)}hrightrvert=0.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 23 hours ago









                              José Carlos Santos

                              139k18111203




                              139k18111203






















                                  up vote
                                  1
                                  down vote













                                  If I am correct then the question mention's about



                                  $$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$



                                  So I would use the following:



                                  $$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$



                                  $$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$



                                  Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.



                                  This shortened the expression to



                                  $$ lim_{h rightarrow 0} g(h) $$



                                  Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.






                                  share|cite|improve this answer

























                                    up vote
                                    1
                                    down vote













                                    If I am correct then the question mention's about



                                    $$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$



                                    So I would use the following:



                                    $$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$



                                    $$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$



                                    Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.



                                    This shortened the expression to



                                    $$ lim_{h rightarrow 0} g(h) $$



                                    Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.






                                    share|cite|improve this answer























                                      up vote
                                      1
                                      down vote










                                      up vote
                                      1
                                      down vote









                                      If I am correct then the question mention's about



                                      $$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$



                                      So I would use the following:



                                      $$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$



                                      $$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$



                                      Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.



                                      This shortened the expression to



                                      $$ lim_{h rightarrow 0} g(h) $$



                                      Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.






                                      share|cite|improve this answer












                                      If I am correct then the question mention's about



                                      $$ f(x) = begin{cases}0 & x = 0\ g(x)sin{x} & x neq0end{cases} $$



                                      So I would use the following:



                                      $$ f^{'}(x) = lim_{h rightarrow 0} frac{f(x+h)-f(x)}{h} $$



                                      $$f^{'}(0) = lim_{h rightarrow 0}frac{f(h)-f(0)}{h} \ = lim_{h rightarrow 0} frac{g(h)sin{h}-0}{h}$$



                                      Now using the identity $ lim_{h rightarrow 0} frac{sin{h}}{h} = 1$.



                                      This shortened the expression to



                                      $$ lim_{h rightarrow 0} g(h) $$



                                      Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 23 hours ago









                                      Lakshya Sinha

                                      464




                                      464






















                                          up vote
                                          0
                                          down vote













                                          By Taylor's theorem:
                                          $$g(x)=g(0)+g'(0)x+r(x)x$$
                                          $$g(x)=xr(x)$$
                                          So
                                          $$f(x)=xr(x)sinleft(frac{1}{x}right)$$
                                          And
                                          $$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
                                          Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.






                                          share|cite|improve this answer

























                                            up vote
                                            0
                                            down vote













                                            By Taylor's theorem:
                                            $$g(x)=g(0)+g'(0)x+r(x)x$$
                                            $$g(x)=xr(x)$$
                                            So
                                            $$f(x)=xr(x)sinleft(frac{1}{x}right)$$
                                            And
                                            $$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
                                            Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.






                                            share|cite|improve this answer























                                              up vote
                                              0
                                              down vote










                                              up vote
                                              0
                                              down vote









                                              By Taylor's theorem:
                                              $$g(x)=g(0)+g'(0)x+r(x)x$$
                                              $$g(x)=xr(x)$$
                                              So
                                              $$f(x)=xr(x)sinleft(frac{1}{x}right)$$
                                              And
                                              $$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
                                              Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.






                                              share|cite|improve this answer












                                              By Taylor's theorem:
                                              $$g(x)=g(0)+g'(0)x+r(x)x$$
                                              $$g(x)=xr(x)$$
                                              So
                                              $$f(x)=xr(x)sinleft(frac{1}{x}right)$$
                                              And
                                              $$f'(0)=lim_{x to 0} frac{f(x)-f(0)}{x}=lim_{x to 0} r(x) sinleft(frac{1}{x}right)=0$$
                                              Because $r(x) to 0$ and $sinleft(frac{1}{x}right)$ is bounded.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered 23 hours ago









                                              Botond

                                              4,9632732




                                              4,9632732






























                                                   

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