(e ^ x ^ 2 + 2cosx-3) / x ^ 4 x tends to 0, and the limit is 7 / 12

(e ^ x ^ 2 + 2cosx-3) / x ^ 4 x tends to 0, and the limit is 7 / 12


Substituting x = 0 to get 0 / 0, lobida, the molecular denominator is (2xe ^ (x ^ 2) - 2sinx) / 4x ^ 30 / 0, lobida (2e ^ (x ^ 2) + 4x ^ 2 * e ^ (x ^ 2) - 2cosx) / 12x ^ 20 / 0, lobida [4xe ^ (x ^ 2) + 8XE ^ (x ^ 2) + 8x ^ 3 * e ^ (x ^ 2) + 2sinx] / 24x = [12xe ^ (x ^ 2) + 8x ^ 3 * e ^ (x ^ 2) + 2sinx] / 24x0 / 0



N tends to infinity. What is the limit of (2 ^ N-N ^ 2) / 3 ^ n? F (x) = root (1-x ^ 2). Find the derivative when x = 2
Hope to understand, do not need to step


N tends to infinity, and the limit of (2 ^ N-N ^ 2) / 3 ^ n is 0
f'(x)=-x/√(1-x^2)
F '(2) meaningless



When x tends to 0, find the limit of x ^ 2 / [1 + x ^ 2 under 1-radical]


The denominator is rationalized by multiplying 1 + root (1 + x ^ 2)
The original formula can be reduced to (1 + x ^ 2) under - 1-radical,
When x approaches 0, the original formula = - 2