Why is $\lim_{x\to0+}x\cot (x)=1$?

Question

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: $\lim\limits_{x\to 0^+}x\cot (x)=0\cdot 0 = 0$?

PS: I know how to find the limit: $\displaystyle\lim_{x\to 0}x\cot x=\lim_{x\to0}\frac {x\cos (x)} {\sin (x)}=\lim_{x\to 0} \cos (x) = 1$ and it's the same with LHR too but I just find it strange since both of them are supposed to be $0$.

Answer 1

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 $0$.

Answer 2

Hint:

$\cot(0)$ isn't equal to $0$, in fact $\cot$ isn't continuous at $x=0$.