Find the limit X - > 0 + (1-cosx) (x ^ x-1) ((√ x + 1) - 1) / (1 + cosx) ^ 2 * LNX * arctan (x ^ 2) ^ 2

Find the limit X - > 0 + (1-cosx) (x ^ x-1) ((√ x + 1) - 1) / (1 + cosx) ^ 2 * LNX * arctan (x ^ 2) ^ 2

(1 - cosx) (x ^ x-1) ((√ x + 1) - 1) / (1 + cosx) ^ 2 * LNX * arctan (x ^ 2) ^ 2 = 1 / (1 + cosx) ^ 2 * (1 - cosx) / x ^ 2 * (x ^ X-1) / xlnx * (√ (x + 1) - 1) / X * [x ^ 2 / arctan (x ^ 2)] ^ 2, when X - > 0 + (1) Lim1 / (1 + cosx) ^ 2 = 1 / 4

What is the limit of [(1 + X * SiNx) ^ 1 / 2 - cosx] / X * SiNx when x tends to 0

Substituting 0 into the known numerator and denominator is 0!
So it is type 0 / 0!
Derivation of numerator and denominator respectively!
Then do it yourself!
Then put 0 into the known numerator, denominator is 0!
Keep on deriving!
Finally, put 0 in!

Given the function f (x), for any real number m, N, f (M + n) = f (m) times f (n), and f (1) = a (a is not equal to 0), f (n)=

Let m = 1
f(n+1)=f(n)*f(1)=a*f(n)
a=f(n+1)/f(n)
therefore
f(n)/f(n-1)=a
f(n-1)/f(n-2)=a
……
f(3)/f(2)=a
f(2)/f(1)=a
Multiply, divide in the middle
f(n)/f(1)=a^(n-1)
f(1)=a
So f (n) = a ^ n