LIM (x → 0 +) lncotx / LNX is used to find the limit

LIM (x → 0 +) lncotx / LNX is used to find the limit

lim(x→0+) lncotx/lnx
=lim(x→0+) (1/cotx)*(-csc^2x)/(1/x)
=-lim(x→0+)x/sinxcosx
=-1

Given the function f (x) = 2ln3x + 8x, then LIM (DX) tends to 0) f (1-2dx) - f (1)
If f (x) = 2ln3x + 8x, then LIM (DX tends to 0)
The value of F (1-2dx) - f (1) / DX is?
DX is delta x, which is triangle X. what's the result? I calculated - 20

f(x) = 2ln(3x) + 8xf(1) = 2ln(3) + 8lim(Δx→0) [f(1 - 2Δx) - f(1)]/Δx= lim(Δx→0) [2ln(3(1 - 2Δx)) + 8(1 - 2Δx) - (2ln(3) + 8)]/Δx= lim(Δx→0) [2ln(3) + 2ln(1 - 2Δx) + 8 - 16Δx - 2ln(3) - 8]...