Question

Suppose f is a function such that
f(x) = x^2 x element Q,
0 x element R − Q.
Is f continuous at c = 0? Is f differentiable at c = 0?

Answer 1

1) The function is continuous at c=0.

Indeed, fix any

epsilon>0
and put
delta = epsilon.

We can assume that
epsilon<1.

Then for any x with
|x|<delta
we have that

|f(x)| < epsilon.

Indeed, if x is an element of Q, then
|f(x)| =
|x^2| < |x| < epsilon, since |x|<1.

if x is an element of R-Q,
then
|f(x)| = 0 < epsilon.

Thus f is continuous at
c=0.


2) f is also differentiable at c=0, in the sense that


lim_{x->0} ( f(x)-f(0) ) / x = 0.

Indeed, fix

epsilon>0
and put
delta = epsilon.

Then for any x with

|x|<delta
we have that
| (f(x)-f(0)) / x | <
epsilon.

Indeed, if x belongs to Q-0, then

| [ f(x)-f(0) ] / x
| = | (x^2 - 0) /x | = |x| < epsilon

and if x belongs to R-Q, then
f(x)=0, and

| (f(x)-f(0)) / x | = | (0 - 0)/x | = 0 <
epsilon

Thus
lim_{x->0} [ f(x)-f(0) ] / x = 0.



3)
On the other hand f is not continuous and not differentiable at any other point
c<>0.
Indeed, if c<>0, then take two sequences
(a_i) and
(b_i) converging to c such that each
a_i is raitonal, i.e. belogns to
Q
and each
b_i is irrational, i.e. belogns to R-Q.

Then

lim_{i -> infinity} f(a_i) = lim_{i -> infinity} a_i^2 =
c^2,
while
lim_{i -> infinity} f(b_i) = lim_{i -> infinity} 0 = 0
<> c^2.

So f is not continuous at c, and therefore it is not
differentaible at c as well.