Jtdylktuy

Prove that the function
$$f(x) = left{
begin{array}{ll}
x^2, & x in mathbb{Q} \
0, & x in mathbb{Q}^c \
end{array}
right.$$

is differentiable at $x = 0.$

I'm not sure how to calculate the limits for this function (the density of $mathbb{Q}$ in $mathbb{R}$ is confusing me). Any advice would be appreciated!

We have
$$lim_{xto 0,xin Bbb {Q}}dfrac{f(x)-f(0)}{x-0}=0$$
and
$$lim_{xto 0,xin mathbb{R}setminusmathbb{Q}}dfrac{f(x)-f(0)}{x-0}=0.$$
Hence, $f$ is differentiable at $x=0$.

We need that

$$limlimits_{x to 0} frac{f(x)}{x} = 0$$ (having already done some simplification, and noting that the derivative must be zero if it exists, since the function is even).

Now, $$frac{f(x)}{x} = left{array{x,&x in mathbb{Q}\0,&x in mathbb{Q}^c}right.$$

Clearly, both the functions $g(x) = x$ and $h(x) = 0$ have the limit $0$ at $0$, so by what I guess you might have called the subset limit theorem, so does $frac{f(x)}{x}$, so the limit is 0. In particular, the limit exists, so $f$ is differentiable at $0$.