calculus - Is $f$ continuous at zero?

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

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 $x=0$?

Answer

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 $0$.