limx→π/6 √3sinx-cosx/x-π/6
Answers
Answered by
4
We establish that the limit
limx→π/61−3–√tanxπ−6x
limx→π/61−3tanxπ−6x
is of the indeterminate form 0000; since numerator and denominator are differentiable, let us attempt De l'Hôpital's rule. It works:
limx→π/61−3–√tanxπ−6x=limx→π/6−3–√1cos2x−6=3–√6⋅43=23–
limx→π/61−3–√tanxπ−6x
limx→π/61−3tanxπ−6x
is of the indeterminate form 0000; since numerator and denominator are differentiable, let us attempt De l'Hôpital's rule. It works:
limx→π/61−3–√tanxπ−6x=limx→π/6−3–√1cos2x−6=3–√6⋅43=23–
Similar questions