Why LIM (x → 0) cosx = 1

Why LIM (x → 0) cosx = 1

Cosx is continuous on R
So LIM (x → 0) cosx = cos0 = 1

Given the function f (x) = (X & # 178; + a) / (x + 1) (a ∈ R) (1), we prove by definition that when a = 3, the function y = f (x) is in [1, positive and negative
(1) It is proved by definition that when a = 3, the function y = f (x) is an increasing function on [1, positive infinity]
(2) If the function y = f (x) has the minimum value - 1 on [1,2], find the value of real number a

(1) When a = 3, f (x) = (X & # 178; + 3) / (x + 1)
Any 1 ≤ x14
∴(x1+1)(x2+1)-4>0
Another x1-x2

LIM ((e ^ x-e ^ (- x)) / (TaNx)) when the limit approaches 0

e^x=1+x;
e^-x=1-x
tanx=x
Limit = 2

Lim [(x-1) / (x + 1)] ^ (x + 2) x tends to infinity, find the limit

[(x-1)/(x+1)]^(x+2) = [1 - 2/(x+1)]^(x+2)
let t=(x+1)/2
[(x-1)/(x+1)]^(x+2) = [1 - 1/t]^(2t+1) = [(1 - 1/t)^t]^2 * (1 - 1/t)
lim (1 - 1/t)^t = 1/e
lim (1 - 1/t) = 1
lim [(x-1)/(x+1)]^(x+2) = 1/e²