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Here the function is $$f(x) =\frac{\sin{x}}{x}.$$ We see that the right hand limit equals the left hand limit but does $f(0)$ exist?

qwr
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Ankan Mandal
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  • It isn't continuous but it can be extended by continuity – Fareed Abi Farraj Apr 04 '19 at 05:23
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    @FareedAF $f:\mathbb{R}-0\to\mathbb{R}$ is continuous, $f:\mathbb{R}\to\mathbb{R}$ is not defined at $0$, so it makes no sense to talk about continuity there. – Maja Blumenstein Apr 04 '19 at 09:46
  • You are right @Maja it makes no sense saying that it is "not continuous" when the function is already "not defined". But it is not wrong to say it, and I did say it in this way to answer his question because he asked "Is it continuous?" The answer is "No it is not". – Fareed Abi Farraj Apr 04 '19 at 12:26

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Indeed, the given function $$f(x) := \frac{\sin{x}}{x}$$ is not defined at $x=0$. Because $f(0)$ isn't defined (at this stage), it does not really make sense to ask if $f$ is continuous at the point $x=0$. This is because continuity at $x=0$ requires that $$ \lim_{x \to 0} f(x) = f(0). $$ However, this is meaningless if we don't know what $f(0)$ is!

Now, as you've observed, the right and left limits agree, i.e. $$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = 1. $$ This means that the function $f$ can be made continuous at $x=0$ by assigning it the value $f(0) := 1$. Alternatively, this means that the function $$ f(x) := \begin{cases} \frac{\sin{x}}{x} & \text{if }x \neq 0,\\ 1 & \text{if }x = 0 \end{cases} $$ is continuous everywhere.

rolandcyp
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As it is $f(0)$ is not defined and $f$ is continuous on $\mathbb R \setminus \{0\}$. If you extend the definition of $f$ by defining $f(0)$ as $1$ then it becomes continuous.

Kavi Rama Murthy
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$f(0)$ is undefined since $\sin(0)/0$ is undefined. However, the limit of $f(x)$ as $x$ approaches zero, as you noted, is one. If you define $f(0)=1$, then $f$ is continuous (everywhere) due to this.

wjmolina
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