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This question already has an answer here:

Is $\dfrac{\sin(x)}{x}$ at $x = 0$ continuous? Whats the value at $x=0~?$

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marked as duplicate by Brevan Ellefsen, Antonios-Alexandros Robotis, Mark Viola, Shailesh, Juniven Apr 13 '17 at 5:54

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  • $\begingroup$ I believe You do know about something Called L'hospital Rule? $\endgroup$ – The Dead Legend Apr 13 '17 at 5:22
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    $\begingroup$ @BrevanEllefsen Why can't you use Hospital? This CAN be approached with Hospital $\endgroup$ – DMH16 Apr 13 '17 at 5:25
  • $\begingroup$ This might be helpful math.stackexchange.com/questions/2092824/… $\endgroup$ – mlc Apr 13 '17 at 5:26
  • $\begingroup$ @DMH16 L'Hopital assumes you know the derivatives of the top and bottom. However, proving that $\frac{d}{dx} \sin(x) = \cos(x)$ usually USES the limit $\lim_{x \to 0} \frac{\sin x}{x}$ and so the logic is circular. depends on how you define the sine function though. Defining it in terms of power functions let's you do this. $\endgroup$ – Brevan Ellefsen Apr 13 '17 at 5:27
  • $\begingroup$ @BrevanEllefsen That is true, however, it wouldn't be wrong to state that the OP can use Hospital to evaluate the limit (if he has been properly introduced to it). You clearly stated that "it doesn't work here", which is incorrect $\endgroup$ – DMH16 Apr 13 '17 at 5:30
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It isn't if you don't define its value at $0$ to be $1$. As it is written, the function $f$, defined by $f(x) = (\sin x)/x$, is not defined at $0$, so it is not continuous at $0$.

However, the function $g$, defined by $$ g(x) = \begin{cases} \dfrac{\sin x}{x} &\text{if $x \neq 0$,}\\[1ex] 1 &\text{if $x = 0$,} \end{cases} $$ is continuous at $0$. Why? Well, this requires you to show that $\lim_{x \to 0} (\sin x)/x = 1$, which I will omit discussing (you can use L'Hospital's rule to handle it, although most calculus textbooks prove it using a geometric argument, since L'Hospital's rule is unavailable at that time to the reader).

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Note that $\sin x = \sum_{k=0}^\infty {x^{2k+1} \over (2k+1) }$, so for $x \neq 0$, we have ${\sin x \over x} = \sum_{k=0}^\infty {x^{2k} \over (2k+1) }$, and since $\sigma(x) = \sum_{k=0}^\infty {x^{2k} \over (2k+1) }$ is analytic everywhere, we see that $\lim_{x \to 0} {\sin x \over x} = \lim_{x \to 0} \sigma(x) = \sigma(0) = 1$.

The function $\sigma$ is also known as $\operatorname{sinc}$.

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