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` .**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now