Let f ( x ) = { x 2 sin ⁡ ( 1 x ) , if x ≠ 0 0 , if x = 0 and g ( x ) = { 0 , if x ∉ Q 1 b , if x = a b with a ∈ Z , b ∈ Z + and gcd ( a , b ) = 1 Let h ( x ) = f ( g ( x ) )Determine where h ( x ) is continuous and differentiable.My try: h ( x ) = { 1 b 2 sin ⁡ ( b ) where x = a b , 0 where x ∉ Q Then we can show h ( x ) is not continuous at every x ∈ Q as following: Assume p = a b then f ( p ) = f ( a b ) = 1 b 2 sin ⁡ ( b ), but given any irrational sequence x n converging we to p have lim n → ∞ f ( x n ) = 0 since x n is irrational, then we proved the discontinuity in Q .But how to prove that h is continuos for x ∉ Q ?

Question

Let      f  (  x  )  =      {                                        x                          2                                sin          ⁡          (                      1            x                    )          ,                                      if                           x              ≠              0                                                            0          ,                                      if                           x              =              0                                              and  g  (  x  )  =      {                            0          ,                          if           x          ∉                      Q                                                            1            b                    ,                          if           x          =                      a            b                                with           a          ∈                      Z                    ,                     b          ∈                                    Z                                      +                                 and           gcd          (          a          ,          b          )          =          1                        Let   h  (  x  )  =  f  (  g  (  x  )  )Determine where   h  (  x  ) is continuous and differentiable.My try:  h  (  x  )  =      {                                        1                          b                              2                                              sin          ⁡          (          b          )                          where           x          =                      a            b                    ,                                      0                          where           x          ∉                      Q                                  Then we can show   h  (  x  ) is not continuous at every   x  ∈      Q   as following: Assume   p  =      a    b   then      f  (  p  )  =  f  (      a    b    )  =      1          b              2              sin  ⁡  (  b  ), but given any irrational sequence       x          n       converging we to   p have       lim          n      →      ∞        f  (      x          n        )  =  0 since       x          n       is irrational, then we proved the discontinuity in       Q  .But how to prove that   h is continuos for   x  ∉      Q  ?

Sharon Dawson · Accepted Answer

To prove   h is continuous for irrational numbers we have to prove that  ∀  x  ∉      Q       ∀  ϵ  &amp;gt;  0     ∃  δ  &amp;gt;  0  :      |    x  −  y      |    &amp;lt;  δ  ⇒  f  (  y  )  &amp;lt;  ϵIf we take an interval of length   1 around   x, we see there are at most   3 fractions that can be written       a    2   in this interval (two such fractions have a distance of       1    2   between them). Similarly,   ∀  n  ∈      N  , there is a finite number of fractions in this interval that can be written       a    n  . But we observe   h  (      a    n    )  ⩽      1          n      2      , which means if we take   n large enough, we can assure   h  (      a    b    )  &amp;lt;  ϵ     ∀  b  &amp;gt;  n. As there are only a finite amount of fractions where the denominator is smaller than   n ( in the chosen interval), there is a minimum distance between   x and such a fraction. If we take   δ smaller than this distance, the condition to be continuous is satisfied.Take any rational point       a    1  .   h  (      a    1    )  ≠  0 (unless       a    1    =  0). We now define the interval       I    1    =  [      a    1    −  h  (      a    1    )  ,      a    1    +  h  (      a    1    )  ]. As       Q   is dense in       R  , we know there is another rational point in       I    1  , call it       a    2  .Now define the interval       I    2    =  [      a    2    −  h  (      a    2    )  ,      a    2    +  h  (      a    2    )  ].The intersection of       I    1   and       I    2   is non-empty, and we find another rational point,       a    3  , in this intersection.We continue, and as the length of these intervals tend to   0, we know       lim          n      →      ∞            a    n   exists. Call it   x. Now   h is clearly not differentiable in   x, as the derivate would be   1 for any       a    1   and   0 for any irrational number. There is no reason   x should be rational, irrational numbers where   h is not differentiable.

Let {x^2 sin(1/x), if x≠0 0, if x=0 and g(x)={0, if x∉Q 1/b, if x=ab with a∈Z, b∈Z+ and gcd(a,b)=1

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