Is $f(y) = \frac{\sin y}{y}$ a continuous function?

Question

I am not sure about the point at $y=0$, the denominator cannot be zero but the numerator is also zero! The limit at $y=0$ exists but what about the original function?

The RHS of the function definition $f(x) = \frac{\sin x}{x}$ is undefined at $x = 0$. However, if you assign a value $1$ to $f(0)$, it will turn it to a continuous function. Since $1$ is the only value which can turn $f(x)$ to a continuous function, when we say $f(x) = \frac{\sin x}{x}$, we usually assume $f(0) = 1$ and work with this extension of $f$ from $\mathbb{R} \setminus {0}$ to $\mathbb{R}$. — achille hui, Apr 05 '20 at 17:09

score 3 · Accepted Answer · answered Apr 05 '20 at 17:02

The definition says that $f : D \to \mathbb{R}$ is continuous if for all $a \in D$ (the domain of $f$),

$$ \lim_{x \to a} f(x) = f(a). $$

So if we're considering continuity at $0$ then first of all $0$ must be in the domain of $f$ and then we're asking if

$$ \lim_{x \to 0} f(x) = f(0). $$

So you see that we need to say what $f(0)$ is before we can make a decision about whether or not $f$ is continuous at $0$.

So there are two reasonable functions we can consider:

$$ f : \mathbb{R} \to \mathbb{R} \text{ defined by } f(x) = \begin{cases} \frac{\sin x}x & x \ne 0 \\ 1 & x = 0 \end{cases} $$

or

$$ f : \mathbb{R} \setminus \{0\} \to \mathbb{R} \text{ defined by } f(x) = \frac{\sin x}x. $$

Both of these functions are continuous. The first is continuous at $0$ because we've defined $f(0)$ as $\lim_{x \to 0} \frac{\sin x}{x}$. The second is also continuous at every point in its domain for the simple reason that $0$ is not in its domain.

Is $f(y) = \frac{\sin y}{y}$ a continuous function?

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