# Why is $\lim\limits_{x\to0+}x\cot x=1$?

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 $0$.

• Actually, $\cot$ is not continuous at $0$. We have $\lim_{x\to 0^+}\cot x = -\lim_{x\to 0^+}\cot x = \infty$ – fonini Feb 12 '15 at 13:40
• @fonini Did you mean $\lim\limits_{x\to 0^+}\cot x = -\lim\limits_{x\to 0^{\color{red}-}}\cot x$? – Workaholic Feb 12 '15 at 13:41
• Your "PS" answer is OK. – zoli Feb 12 '15 at 13:51
• yes, of course, sorry – fonini Feb 12 '15 at 13:51

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 $0$.

Hint:

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

• Ohh right.. I mixed it up with an upside down $\tan x$... – GinKin Feb 12 '15 at 13:41