Question

Calculate the limit of the function with L'Hospital rule​

Answer 1

L=24

Explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if
`\lim_(x \to a) (f(x))/(g(x))` takes
`(0)/(0)` or
`(\infty)/(\infty)` form, then,

`\lim_(x \to a) (f(x))/(g(x))= \lim_(x \to a) (f'(x))/(g'(x))`

where
`f'(x)=(df(x))/(dx)` and
`g'(x)=(dg(x))/(dx)`

Now coming to the problem,

`L= \lim_{x \to (\pi)/(6) } (cot^(3)x-3cotx)/(cos(x+(\pi)/(3) ))`

Here
`f(x)=cot^(3)x-3cotx` and
`g(x)=cos(x+(\pi)/(3) )`

Substituting
`x=(\pi)/(6)` in f(x) and g(x),

`f((\pi)/(6))=cot^(3)(\pi)/(6)-3cot(\pi)/(6)\\=3√(3)-3√(3)\\ =0`

`g((\pi)/(6))=cos((\pi)/(6)+(\pi)/(3))\\=cos(\pi)/(2)\\=0`

Since L takes the form
`(0)/(0)`, using l'hopital's rule

`L= \lim_{x \to (\pi)/(6)} (cot^(3)x-3cotx)/(cos(x+(\pi)/(3)))= \lim_{x \to (\pi)/(6)} (3cot^(2)x(-cosec^(2)x)-3(-cosec^(2)x))/(-sin(x+(\pi)/(3)))`

now substituting
`x=(\pi)/(6)` ,

`L= \lim_{x \to (\pi)/(6)} (3cot^(2)(\pi)/(6)(-cosec^(2)(\pi)/(6))-3(-cosec^(2)(\pi)/(6)))/(-sin((\pi)/(6)+(\pi)/(3)))\\=(3*3^(2)(-2^(2))+3(2^(2)))/(-1)\\=24`