lim x→π[(√1-tanx)-(√1+tanx)]/sin2x

lim x→π[(√1-tanx)-(√1+tanx)]/sin2x

The physicochemical Lim x → π [(√ 1-tanx) - (√ 1 + TaNx)] [(√ 1-tanx) + (√ 1 + TaNx)] / ([(√ 1-tanx) + (√ 1 + TaNx)] sin2x) = Lim x → π (- 2tanx) / (((√ 1-tanx) + (√ 1 + TaNx)] sin2x is divided into two limits = Lim x → π (- 2tanx) / sin2x * Lim x → π 1 / [(√ 1 -

LIM (x tends to zero), what is Tan 6x / sin 2x
LIM (x tends to 1) how much is the sixth power of x minus 1, divided by the tenth power of x minus 1
But. I still don't understand my second question. Why is it that after deriving, there is no one? I'm not good at basic skills. I forgot what I learned before.
In addition, is there any way to do this problem? If you don't need to seek derivatives, how can you solve it
What's the formula of Tan's problem? Otherwise, how can you think of this solution of multiplying 6 / 2?

LIM (x tends to 0), tan6x / sin2x = LIM (x tends to 0), [(tan6x / 6x) / (sin2x / 2x)] * (6 / 2) = 1 × 1 × 6 / 2 = 3lim (x tends to 1) (x ^ 6-1) / (x ^ 10-1) = LIM (x tends to 1) (x ^ 6-1) '/ (x ^ 10-1)' = LIM (x tends to 1) 6x ^ 5 / 10x ^ 9 = 6 / 10 = 3 / 5

Find Lim x tends to 0 (sin2x) / x?

When the important limit x tends to 0, SiNx / X tends to 1?
In the same way, (sin2x) / 2x tends to 1
So (sin2x) / X tends to 2
Actually, remember,
When x tends to zero, SiNx is equivalent to X
So sin2x is equivalent to 2x
Of course, (sin2x) / X is equivalent to 2x / x, so the limit value is 2