Question #caf18

Question

248 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-05T12:47:54

See below.

#cot(x/2) = cos(x/2)/sin(x/2)#

We can reason that when #x-> pi# , #sin(x/2)=1#. This is then an irrelevant factor.

So this leaves us:

#cos(x/2)/(pi-x)#

Using L'Hospital's Rule:

#d/dx(cos(x/2))=-1/2sin(x/2)#

#d/dx(pi-x)=-1#

#:.#

#(-1/2sin(x/2))/-1#

Plugging in #x=pi#

#(-1/2sin(pi/2))/-1=(-1/2)/-1=1/2#

#:,#

#lim_(x->pi)(cot(x/2))/(pi-x)=1/2#