Why is $\lim\limits_{x\to0+}x\cot x=1$?
Since both $x$ and $\cot x$ are continuous at zero and both equal to zero at $x=0$ why is the limit of both of them $1$?
i.e why isn't it: $\lim\limits_{x\to0+}x\cot x=0\cdot 0 = 0$?
PS: I know how to find the limit: $\displaystyle\lim_{x\to0}x\cot x=\lim_{x\to0}\frac {x\cos x} {\sin x}=\lim_{x\to0} \cos x = 1$ and it's the same with LHR too but I just find it strange since both of them are supposed to be $0$.
Answer
The cotangent is the reciprocal (the multiplicative inverse) of the tangent, that is $1/ \tan x$. The tangent is $0$ at $0$ so its reciprocal has a pole at $0$.
It is important to note that while the cotangent is $(\tan x)^{-1}$ this is not the same as $\tan^{-1} (x)$, the inverse function (the functional inverse) of the tangent also called arcus tangent. This would be $0$ at $0$.
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