# Limits Question Limit as h approaches 0 (cot(5pi/6 +h)-cot(5pi/6))/h

You need to substitute 0 for h in equation under limit such that:

`lim_(h-gt0) (cot(5pi/6 + h) - cot (5pi/6))/h = (cot(5pi/6) - cot (5pi/6))/0 = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(h-gt0) ((cot(5pi/6 + h) - cot (5pi/6))')/(h') = lim_(h-gt0) -(((5pi/6 + h)')/(sin^2(5pi/6+h)))/1`

`lim_(h-gt0) -(((5pi/6 + h)')/(sin^2(5pi/6+h)))/1 = lim_(h-gt0) -1/(sin^2(5pi/6+h))`

You need to substitute 0 for h in equation under limit such that:

`lim_(h-gt0) -1/(sin^2(5pi/6+h)) = -1/(sin^2(5pi/6)) = -1/(sin^2(pi/6))`

`lim_(h-gt0) -1/(sin^2(5pi/6+h)) = -1/((1/2)^2) = -4`

Hence, evaluating the limit to the function when h approaches to 0 yields `lim_(h-gt0) (cot(5pi/6 + h) - cot (5pi/6))/h = -4` .

Approved by eNotes Editorial Team