Let $n$ be a positive integer and let
$$f(x) = x^{2n} \cdot \sin\left(\frac{1}{x}\right)$$
$$f(0) = 0$$
Then mathematical induction can be used to prove: (a) The $n$th derivative of $f(x)$ exists for each value of $x$. (b) The $n$th derivative $f(x)$ is not continuous at $x = 0$.
Let $n$ be a positive integer and let
$$g(x) = x^{2n+1} \cdot \sin\left(\frac{1}{x}\right)$$
$$g(0) = 0$$
Then mathematical induction can be used to prove: (a) The $n$th derivative of $g(x)$ is continuous at each value of $x$. (b) The $(n+1)$st derivative of $g(x)$ does not exist at $x = 0$.
More generally, let $a$, $b$ be positive real numbers, let $n$ be a positive integer, and define $h(x)$ by:
$$h(x) = x^a \cdot \sin\left(\frac{1}{x^b}\right)$$
$$h(0) = 0$$
The $n$th derivative of $h(x)$ exists for all values of $x$ if and only if $a > n + (n-1)b$.
The $n$th derivative of $h(x)$ is bounded on every bounded interval if and only if $a \geq n + nb$.
The $n$th derivative of $h(x)$ is continuous at each point if and only if $a > n + nb$.
To prove these statements, you can use mathematical induction to prove that the $n$th derivative of $h(x)$ has the form
$$P_{n}(x) \cdot \cos\left(\frac{1}{x^b}\right) + Q_{n}(x) \cdot \sin\left(\frac{1}{x^b}\right),$$
where $P_n$ and $Q_n$ are polynomials such that at least one of them has a lowest degree term that is a NONZERO multiple of $x^{a-n-nb}$ and neither has a lower degree term. [The added emphasis on nonzero is because this becomes vital for the "only if" halves of the statements above.] To prove the "if" halves, you may want to incorporate into the induction statement the fact that the $n$th derivative at $x = 0$ is zero.