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I'm following the MIT OpenCourseware course on calculus, and when proving the derivative of $\sin(x)$ the following assumption is needed

$\lim\limits_{x\to0} \frac{\sin(x)}{x} = 1$

The proof for this is given and I can follow it.

My question however relates to my initial intuition. Which is that the limit of $\sin(x)$ at 0 is defined: $\sin(0) = 0$. So Why is it wrong to just plug that in

$ \frac{\sin(0)}{0} = 0$

I would have never looked further because this seems like a simple problem. What is wrong with my intuition here?

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I had thought this was a duplicate, but the issue might really be about $\frac{0}{0}$, not $\frac{\sin(x)}{x}$. – robjohn Dec 23 '11 at 14:20
Intuition should always be driven by computation, not by appearances. If you had ever calculated, for instance, $\sin(.01)/.01$, you would never have gotten that “intuition”. – Lubin Dec 23 '11 at 19:56
Your intuition was based on a single point (x = 0). Limits are based on the behaviour in an area around the point. sin (0.1) / 0.1 is quite close to 1 already. Your intuition was also wrong, because 0/0 isn't zero, it's actually undefined, so your intuition should have been "this might be tricky and needs a closer look". – gnasher729 May 14 '14 at 8:02
up vote 13 down vote accepted

You are making two fundamental mistakes:

First: after substituting $x=0$, you wind up dividing by $0$.
This is an undefined expression; so, you should suspect that things have gone awry. In your example, an even worse thing happens: you wind up with $0\over0$; which (if it makes any sense) is even more "undefined" than, say, $1\over 0$.

For instance $\lim\limits_{x\rightarrow 0}{2x\over x}=2$ and $\lim\limits_{x\rightarrow 0}{ x\over 2x}={1\over2}$.

In fact a 0/0 form could possibly be anything.

Secondly, when taking limits, you should really avoid just plugging in the limit point into the expression. This only works in certain cases (when the limit exists and the function is continuous at the limit point), and, more importantly, hides the true nature of what a limit is:

Informally, you write $$\lim_{x \rightarrow a} f(x)=L$$ if the values of $f(x)$ get closer and closer to $L$ as $x$ gets closer and closer to, but not equal to, $a$.

Note the phrase "not equal to" in the above. This is an important restriction and is the reason why just substituting the limit point into what you're taking the limit of can lead to trouble.

As an example of determining the limit of a 0/0 form, let's try to find
$$ \tag{1} \lim_{x\rightarrow 1}{(x+1)(x-1)\over (x-1)} $$

Formally substitutiing $x=1$ in the above gives $0/0$.

If, however, $x$ is close to one, but different from 1, then substitution into the right hand side of (1) gives a well-defined number; moreover, the $(x-1)$ terms will cancel and the final result will be just $x+1$.
Thus, if $x$ is close to 1, but different from 1, the expression ${(x+1)(x-1)\over (x-1)}$ is close to 2.

As it turns out, the closer $x$ is to 1, but different from 1, the closer ${(x+1)(x-1)\over (x-1)}$ is to 2; and thus $\lim\limits_{x\rightarrow 1}{(x+1)(x-1)\over (x-1)}=2$.

Finding your limit ($\thinspace\lim\limits_{x\rightarrow0}{\sin x\over x} \thinspace$) is a bit more complicated than the above. You can find the standard derivation of this limit, which utilizes the Squeeze Theorem
and simple geometry here (or at your course site, as you said).

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You can just use L'Hôpital's rule.

$$\begin{align*} \frac{d}{dx} \sin(x) &=\cos(x) \\ \frac{d}{dx} x &= 1 \\ \lim_{x\rightarrow0} \frac{\sin(x)}{x}&=\lim_{x\rightarrow0}\frac{\cos(x)}{1}=1 \end{align*}$$

Edit: I think this rule is very helpful to evaluate limits involving indeterminate forms. Not as proof but as a sanity check of an intuitive answer.

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While this is formally correct, how do you know that derivative of sine is cosine? The normal derivation of this fact relies on knowing the limit $\frac{\sin h}{h}$ already. So, to avoid circular reasoning, it has to be proved another way. (Also, you did not answer the question why plugging 0 is incorrect.) – sdcvvc Dec 23 '11 at 15:39
@sdcvvc read my edit. I think it's a quick and helpful way of putting intuition to the test. – Flying Swissman Dec 23 '11 at 15:46
Yes, but this limit is (supposed to be) taught way before someone is taught how to differentiate trigonometric functions... – J. M. Dec 23 '11 at 15:48
Not only that L'H would be circular reasoning, but that limit is actually the definition of the derivative of sin: $\lim_{x\rightarrow0} \frac{sin(x)}{x}=\lim_{x\rightarrow0} \frac{\sin(x)-\sin(0)}{x-0}$... So using the derivative to calculate it is wrong on many levels... – N. S. Dec 23 '11 at 16:54
+1. This question is about intuition. When you're supposed to quickly tell the result of some expression, it doesn't matter if there's circular reasoning involved, it only matters that you start your reasoning from something you know to be true – and $\tfrac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\cos(x)$ is a far easier (and in general more useful) identity to memorize than that limit. What's more, @N.S., it is possible to define the trig functions by their derivative (as a solution to a system of differential equations), in fact I consider this the best definition. – leftaroundabout Dec 23 '11 at 18:58

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