How to find the limit of limsin (x - π / 3) / 1-2cosx when x tends to π / 3? Note, not with the law of Robida and the like, but with the exchange method

How to find the limit of limsin (x - π / 3) / 1-2cosx when x tends to π / 3? Note, not with the law of Robida and the like, but with the exchange method

Sorry, I misread the denominator. Yes, it's a breakpoint
Let X - π / 3 = t, then x = π / 3 + T, X → π / 3 is equivalent to t → 0
The original limit = LIM (t → 0) Sint / [1-2cos (π / 3 + T)]
Molecule = sint
The denominator is expanded by the sum angle formula of cosine, and the denominator is 1-2 (1 / 2 · cost - 3 / 2 · Sint) = 1-cost + 3 · Sint
Since you said that you have not yet learned the law of lobita, you can only consider four operations of limit after you have done this step
If the numerator and denominator are divided by Sint at the same time, the numerator becomes 1 and the denominator becomes (1-cost) / sint ＋ 3
In this case, we just need to find the limit of (1-cost) / Sint,
LIM (t → 0) (1-cost) / Sint = 0.5T & # 178; / T = 0.5T = 0.5T
(above is the Equivalent Infinitesimal Substitution of 1-cost and Sint, 1-cost 0.5 T & # 178;, Sint T)
So the original limit = 1 / (0 ＋ 3) = 1 / √ 3
If you say you haven't learned the replacement of "equivalent infinitesimal", then there is no way. It should be noted that after expanding the denominator, you can't immediately replace it with equivalent infinitesimal, because the replacement theorem of equivalent infinitesimal only allows the replacement of "multiplication and division factor" or the whole replacement of numerator and denominator, and does not allow the replacement of a certain item of addition, even if the final result is the same, It will also be considered wrong