Is $\sin(x)/ \tan(x) = \cos(x)$ at $0$?

We know that $\frac{\sin(x)}{\tan(x)} = \cos(x)$. But at $x = 0$, the LHS becomes $0/0$. So is the function undefined at that point?

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Thus we say that $\frac{\sin\,x}{\tan\,x}$ has a removable discontinuity at $x=0$... –  Ｊ. Ｍ. Sep 1 '11 at 12:34
@brita: that's how we evaluate the limit. For a function to be continuous, the limit has to agree with the defined value, which we don't have (but could create by assigning it the value of 1) –  Altar Ego Sep 1 '11 at 12:44
What @J.M. means is that the function as written is technically not defined at $0$ since plugging in $0$ gives you the meaningless expression $0/0$. But, as you point out, the function has a definite limit ($=1$) as $x \to 0$. So, one could make it continuous by redefining its value at $0$ to be $1$. This is why we say that the function has a removable discontinuity there. –  Srivatsan Sep 1 '11 at 12:45
Ok I understand in the situation for 0/0. But what about 1/(1/x)=x? Is the LHS also undefined at x=0? Ie. is there also a "hole" at 0? –  brita Sep 1 '11 at 12:55
Ok, I think I'm confused with the definition of "undefined". So 0/0, 1/0, 1/$\infty$ are all undefined, is that right? If I were to plot the graph of 1/(1/x) and x side by side, the graphs would be identical except at 0, where the graph of 1/(1/x) would have a "hole". Is that correct? –  brita Sep 1 '11 at 13:09

The general convention for real-valued functions of real variable (at least as discussed in most calculus books) is:

If a function $f$ is given by a formula, and no domain is explicitly specified, then the domain of $f$ is understood to be the natural domain, that is, the set of all $x$ for which the formula "makes sense."

And generally, for two functions to be considered equal you need them to at least have the same domain and the same value at every point of the domain (whether the codomain matters or not is a matter of context and definitions).

From that point of view, the functions $$f(x) = \frac{\sin x}{\tan x}\quad\text{and}\quad g(x) = \cos x$$ are not "the same function": the natural domain of $f(x)$ consists of all $x$ that are not integer multiples of $\pi/2$, whereas the natural domain of $g(x)$ is all real numbers. On those numbers where they are both defined, the functions agree.

So, tecnically, these two functions are not the same function.

Sometimes, we want to relax the condition on the domain a bit and consider the "removable discontinuities." Namely, given a function $f(x)$, if $x=a$ is a point that is not on the domain of $f(x)$, but where $\lim\limits_{x\to a}f(x)$ exists, then we "redefine" $f(x)$ to have this limit as its value at $x=a$, and include $x=a$ into the domain. Here, because $f(x)$ and $g(x)$ agree everywhere except at the integer multiples of $\pi/2$, where $f(x)$ is not defined, we sometimes fudge the difference. But, formally speaking, when we write $$\frac{\sin x}{\tan x} = \cos x,$$ we really mean "wherever they are both defined", just like when we write $$\sec^2 x - \tan^2 x = 1$$ we are saying that this equality holds at the points where the left hand side is defined, and are making no assertion about what happens at the points outside the domains of $\sec(x)$ and $\tan(x)$.

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The general convention for complex-valued functions of complex variable is: remove the removable discontinuities, e.g. $$\lim_{x\to 0} \frac{\sin x}{\tan x} = 1$$ so take that to be the value of the function at 0. (Except that it's conventional to use $z$ rather than $x$ here, and often one uses $x$ and $y$ for the real and imaginary parts of $z$, i.e. $z = x+iy$ and $x$ and $y$ are real.)
But could I add this: It seems to me that that that same convention frequently makes sense in trigonometry with no complex variables and no calculus to be seen anywhere. And in trigonometry, it also usually makes sense to follow the same convention as in complex variables, that there is only one $\infty$, not a $+\infty$ and a $-\infty$. That makes the tangent, cotangent, secant, and cosecant functions continuous everywhere.