The limit of (LN Tan 7x) / (LN Tan 2x) when x approaches 0 Why not directly (1 / tan7x) / (1 / tan2x) = tan2x / tan7x = 2 / 7?

The limit of (LN Tan 7x) / (LN Tan 2x) when x approaches 0 Why not directly (1 / tan7x) / (1 / tan2x) = tan2x / tan7x = 2 / 7?

Because when you use the lobita rule, you're wrong
(ln tan7x)'
=(1/tan7x)*(tan7x)'
=(1/tan7x)*(sec²7x)*(7x)'
=7(1/tan7x)*(sec²7x)
=7/(sin7xcos7x)
In the same way
(ln tan4x)'=4/(sin4xcos4x)

The limit of sin (SiNx) / x, X tends to 0

sin(sin(x))/x
=[sin(sin(x))/sin(x)]*sin(x)/x
=1*1
=1

The limit of sin (SiNx) / x, X tending to 0
sin(sin(x))/x
=[sin(sin(x))/sin(x)]*sin(x)/x
=1*1
=1
I don't understand sin (sin (x)) / sin (x)

[sin(sin(x))/sin(x)] * 【sin(x)/x】
Don't you see a sin (x) is added to the following numerator, which just eliminates it~
In this way, we can make up two pairs of sin () / () forms
Use the law of Robita again = 1