What is the limit of $f(x)$ as $x \to 0$ if it exists? Justify your answers. $$f(x)= \begin{cases} \sin(x) & \text{if }x\notin\mathbb{Q} \\0 & \text{if }x\in \mathbb{Q} \end{cases}$$
What I did: Define two sequences $a_n=\frac{1}{n}$ and $b_n= 2\pi n -\frac{3\pi}{2}$.
Then $f(a_n) \to 0$ as $n \to \infty $ and $f(b_n) \to 1$ as $n \to \infty $. Hence this limit does not exist.
The answer is that you are not choosing a correct pair of sequences to prove that the limit doesn't exist, since the limit actually exists! It is equal to $0$. This follows from the squeeze theorem and the fact that$$(\forall x\in\mathbb{R}):\bigl\lvert f(x)\bigr\rvert\leqslant|x|.$$