$$\lim_{x\to0} \frac{\sin x}x=1 $$

but can we say $\frac {\sin x}x$ is 1 at $x=0$ (exact at point $0$). I think that's absurd as it is undefined there.

Please help me if I am wrong.

  • 2
    $\begingroup$ No, you can't say that, because $x=0$ is not allowed. What you can do is redefine your function and say that it is 1 for $x=0$ and $\sin(x)/x$ for all other values. $\endgroup$ – MrYouMath Apr 8 '16 at 8:26
  • $\begingroup$ But in practice, where there is just one undefined point and it is easily filled in as a limit, we might say that $(\sin x)/x=1$ at $x=0$, because it is obvious what is going on. It would be different if there was a discontinuity at 0, so that the left limit did not equal the right limit. $\endgroup$ – almagest Apr 8 '16 at 8:43
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    $\begingroup$ @almagest: you might say that. I wouldn't! $\endgroup$ – TonyK Apr 8 '16 at 8:52
  • $\begingroup$ what is true is that the function $\text{sinc}(x) = \frac{\sin x}{x}$ for $x \ne 0$ is $C^\infty$ even at $0$ after prolongating it (and its derivatives) by continuity. $\endgroup$ – reuns Apr 8 '16 at 10:54

The $\mbox{sinc}$ function is defined to be:

$$ \mbox{sinc}(x)= \left\{ \begin{array}{ll} \frac{\sin x}{x}, & x\ne 0 \\ 1, & x=0 \\ \end{array} \right. $$

for which we can say $\mbox{sinc}(0)=1$, though $\frac{\sin x}{x}$ is undefined at $x=0$

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