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This question already has an answer here:

I am looking for some basic intuition for $$\lim_{x \to 0}\frac{\sin x}{x}$$ If we look at it as $f(x)=\sin x\cdot\frac{1}{x}$ so as ${x \to 0}$ $\sin x \to 1$ and $\frac{1}{x} \to \infty $ so it is $0 \cdot \infty$ so why is it equals 1?

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marked as duplicate by Ali Caglayan, Venus, Eric Stucky, Claude Leibovici, Ivo Terek Jan 15 '15 at 9:08

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  • $\begingroup$ First of all you have a typo: $\sin x\to 0$, not to 1. The intuition is anyway that for $x\to0$ $x$ and the sine are very close, almost equal, hence the 1. $\endgroup$ – MickG Jan 15 '15 at 7:46
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    $\begingroup$ If you look at the graphs of $\sin x$ and $x$, you can see that around $0$ they "join up", and they do so faster than any of them joins up with the $x$-axis. That is why their ratio goes to $1$. $\endgroup$ – Arthur Jan 15 '15 at 7:51
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    $\begingroup$ There are several geometric ways. Draw the unit circle, centre $O$. Let $A$ and $B$ be on the circle, very close to each other, with $\angle AOB=2x$ (radians). Then $\triangle AOB$ has area $\sin x$, and sector $AOB$ has area $x$. For tiny $x$, the ratio of these areas is nearly $1$. $\endgroup$ – André Nicolas Jan 15 '15 at 7:54
  • $\begingroup$ Notice that $sin\left(x\right)$ can be closely approximated by $x$ for values close to $x=0$ on both sides of the origin. Therefore, as we approach the origin, the graph of that limit is indeed 1. If you have taken calc II you can look at the series expansion of $\sin\left(x\right)$ and you'll find after dividing every term by $x$ and then allowing $x\rightarrow 0$ you're left with 1 as well. $\endgroup$ – jm324354 Jan 15 '15 at 7:54
  • $\begingroup$ This question is definitely not a duplicate of that question, but I think it was reasonably answered by the diagrams in that question, so I also voted to close. $\endgroup$ – Eric Stucky Jan 15 '15 at 8:21
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For $x \in (0,\pi/2) \Rightarrow \cos x < \dfrac{\sin x}{x} < 1$, and use Squeeze Lemma to get the limit of $1$. For $x < 0$, put $x' = - x >0$, and repeat Squeeze Lemma again.

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For $x$ near zero we know that, $\sin x \sim x$. So,

$$\lim_{x \to 0} \dfrac{\sin x}{x} \sim \lim_{x \to 0} \dfrac{x}{x} = 1.$$

To verify, analytically, one application of L'Hospital's rule should do the trick by transforming $$\lim_{x \to 0} \dfrac{\sin x}{x}$$ to $$\lim_{x \to 0} \dfrac{\cos x}{1} = 1.$$

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  • $\begingroup$ Careful with L'Hospital here. You need to know this limit value in order to calculate the derivative of $\sin x$ in the first place. $\endgroup$ – Arthur Jan 15 '15 at 7:53
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Once you know a bit about power series this limit becomes very easy to compute, because $\sin x = x - x^3/6 + x^5R(x)$, so $\sin x / x = 1-x^2 + x^4R(x)$, where $R(x)$ is so small, that we don't care about it and hence the limit is equal to $1$.

If you want to know the whole power series for sine, here it is: $$\sin x = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k+1)!}$$

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  • $\begingroup$ Careful using power series. If your power series is derived by differentiating $\sin x$, then this is a circular argument, since you need to know $\lim_{x\to0}\frac{\sin x}{x}$ in order to calculate the derivative of $\sin x$. $\endgroup$ – Arthur Jan 15 '15 at 7:55
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    $\begingroup$ It all depends on your definition of sine. Rudin defines $\sin t$ as the imaginary part of $\exp(it)$ and derives the power series without calculating limit of $\sin x / x$. $\endgroup$ – user207868 Jan 15 '15 at 8:00

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