Is $f$ continuous at zero? $$\require{cancel}$$

$$f(x) =
\begin{cases}
\frac{\sin x}{|x|} &\text{ if }x \neq0
\\
\hspace{0.3cm}1 &\text{ if }x=0.
\end{cases}$$

My Attempt
1)$$\lim_{x\rightarrow0}\frac{\sin x}{|x|} = \lim_{x\rightarrow0}\frac{\sin x}{x} \frac{x}{|x|} = 1\lim_{x \rightarrow0}\frac{x}{|x|} 
\\$$
2)$$\lim_{x\rightarrow0^{-}}\frac{x}{|x|}=-1 \hspace{0.3cm}\text{and}\hspace{0.3cm} \lim_{x\rightarrow0^{+}}\frac{x}{|x|}=1
$$
Therefore:
$$1\lim_{x\rightarrow0}\frac{x}{|x|}=DNE
$$
so, $f$ is not continuous at $0$.
My question is does my solution actually prove that $f$ is not continuous at $0$? or is it continuous at zero because $f(x)=1$ when $x=0$?


1)$$\lim_{x\rightarrow0}\frac{\sin x}{|x|} = \lim_{x\rightarrow0}\frac{\sin x}{x} \frac{x}{|x|} = 1\lim_{x \rightarrow0}\frac{x}{|x|} 
\\$$

Note that the second equality does not hold, since for two sequences $(a_n)$, $(b_n)$ you only have
$$
\lim_{n \to \infty} a_n b_n = \lim_{n \to \infty} a_n \lim_{n \to \infty} b_n
$$
provided that both sequences converge. Hence in your case it would be better to start directly with a modification of part 2) :
If $f(x)$ is continuous at $0$ then the following equality is necessary:
$$
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x).
$$
But on the one hand you have
$$
\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin(x)}{|x|} = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1
$$
and on the other hand
$$
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(x)}{|x|} = \lim_{x \to 0^-} \frac{\sin(x)}{-x}  = -\lim_{x \to 0^+} \frac{\sin(x)}{x} = -1.
$$
So $f$ can't be continuous at $0$.
