# Why is cot(x) continious in (0,pi) ?

why is cot(x) continious in (0,pi) ???

why is cot(x) continious in (0,pi) ???
I mean Cot(x)=1/Tan(x) , now at pi/2 , tan(x) tends to infinity => 1/tan(x) tends to 0 , now
1/tan(x) is certainly not = 0 , therefore how can cot(x) be continious ?????????

Integral
Staff Emeritus
Gold Member

So for values near, but less then pi/2, Tan goes to +infinity; for values near, but greater then pi/2, tan goes to -infinity. What does that mean about the continuity of cot?

In a case like this you need to look at the two limits of 1/tan, one coming from the left, the other the right. If they are equal then cot is continuous. In this case both approach zero, so the function is continuous and has the value zero.

So for values near, but less then pi/2, Tan goes to +infinity; for values near, but greater then pi/2, tan goes to -infinity. What does that mean about the continuity of cot?

In a case like this you need to look at the two limits of 1/tan, one coming from the left, the other the right. If they are equal then cot is continuous. In this case both approach zero, so the function is continuous and has the value zero.

for a function to be continious at c LHL= RHL at c and also lim at c = f(c) , but here cot(pi/2) does not exist !!

Integral
Staff Emeritus
Gold Member

How is it not zero?

How is it not zero?

cot(pi/2) only tends to 0 , but is never 0!!!! by defination of limit .

HallsofIvy
Homework Helper

No, that does not follow and limits have nothing to do with it. cot(x) is defined as "cos(x)/sin(x)". When x= $\pi/2$, $cos(\pi/2)= 0$ and $sin(\pi/2= 1$ so $cot(\pi/2)= 0$.

You seem to be thinking that cot(x)= 1/tan(x) for all x. It isn't- that is only true as long as both cot(x) and tan(x) exist.