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 $0$.

  • $\begingroup$ Actually, $\cot$ is not continuous at $0$. We have $\lim_{x\to 0^+}\cot x = -\lim_{x\to 0^+}\cot x = \infty$ $\endgroup$ – fonini Feb 12 '15 at 13:40
  • 1
    $\begingroup$ @fonini Did you mean $\lim\limits_{x\to 0^+}\cot x = -\lim\limits_{x\to 0^{\color{red}-}}\cot x$? $\endgroup$ – Workaholic Feb 12 '15 at 13:41
  • 1
    $\begingroup$ Your "PS" answer is OK. $\endgroup$ – zoli Feb 12 '15 at 13:51
  • $\begingroup$ yes, of course, sorry $\endgroup$ – 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 $0$.



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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.