Solving problems with two important limits lim 2x+3 ( ——— ) x →∞ 2x+1

Solving problems with two important limits lim 2x+3 ( ——— ) x →∞ 2x+1

1, LIM (1 + 2 / 2x + 1) ^ (2x + 1) / 2 * 2 / (2x + 1) = lime ^ 0 = 1, the best way is to divide x directly up and down, and then the limit of 1 / X is 0

Two questions about the limit of higher numbers
Sinlix-1
x->0
2 lim (1/x-1/ex-1)
x->0
(the formula is not easy to type, ex is the x power of E)

Both problems are solved by the law of lobita
lim （ x-sinx）/x^2
=lim(1-cosx)/2x
=lim-sinx/2
=0
lim[(1/x-1/(ex-1)]
=Lim {[(EX-1) - x] / [x (EX-1)]} general division
=LIM (e ^ x-1) / (e ^ X-1 + Xe ^ x) lobida
=Lime ^ X / (e ^ x + e ^ x + Xe ^ x) lobida
=1/（1+1+0）
=1/2

Find the following limit in the high number - 2.6
1) LIM (x tends to ∞) (1 + K / x) ^ x (K ≠ 0, integer)
Let y = K / x, then x = K / y, X tends to ∞, which is equivalent to y tends to 0
LIM (x tends to ∞) (1 + K / x) ^ x = LIM (y tends to 0) (1 + y) ^ k / Y
=LIM (y tends to 0) [(1 + y) ^ (1 / y)] ^ k
=[LIM (y tends to 0) [(1 + y) ^ (1 / y)]
=e^k
What I don't understand is why
LIM (y tends to 0) [(1 + y) ^ (1 / y)] ^ k = [LIM (y tends to 0) (1 + y) ^ (1 / y)] ^ y
The k-th power in the equation can be moved to the left of the LIM bracket. Why?

LIM (y tends to 0) [(1 + y) ^ (1 / y)] ^ k = [LIM (y tends to 0) (1 + y) ^ (1 / y)] ^ k
K is a constant,
The 0.9 cycle tends to 1. The 10 times of your 0.9 cycle is the same as the 10 times of 1