Paano ko malulutas ang limitasyong ito?

Sagot:

# e ^ a * (a / 2) * (1 - a) #

Paliwanag:

# "Maaari mong gamitin ang Taylor serye at i-drop mas mataas na mga tuntunin ng order sa" # # "limitasyon para sa" x-> 0 "." #

# x ^ y = exp (y * ln (x)) #

# => (1 + x) ^ y = exp (y * ln (1 + x)) #

# "at" ln (1 + x) = x - x ^ 2/2 + x ^ 3/3 - … #

# "at" exp (x) = 1 + x + x ^ 2/2 + x ^ 3/6 + x ^ 4/24 + … #

# "So" #

#exp (y * ln (1 + x)) = exp (y * (x - x ^ 2/2 + …)) #

# => (1 + x) ^ (a / x) = exp ((a / x) * ln (1 + x)) #

# = exp ((a / x) * (x - x ^ 2/2 + x ^ 3/3 - …)) #

# = exp (a - a * x / 2 + a * x ^ 2/3 - …) #

# => (1 + palakol) ^ (1 / x) = exp ((1 / x) * ln (1 + palakol)) #

# = exp ((1 / x) * (palakol - (palakol) ^ 2/2 + (palakol) ^ 3/3 - …)) #

# = exp (a - a ^ 2 * x / 2 + a ^ 3 * x ^ 2/3 - …) #

# => (1 + palakol) ^ (1 / x) - (1 + x) ^ (a / x) #

# ~~ exp (a - a ^ 2 * x / 2 + …) - exp (a - a * x / 2 + …) #

# ~~ exp (a) / exp (a ^ 2 * x / 2) - exp (a) / exp (a * x / 2) #

# = exp (a) (exp (-a ^ 2 * x / 2) - exp (-a * x / 2)) #

# ~~ exp (a) (1 - a ^ 2 * x / 2 - 1 + a * x / 2) #

# = exp (a) ((x / 2) (a - a ^ 2)) #

# => ((1 + palakol) ^ (1 / x) - (1 + x) ^ (a / x)) / x #

# ~ ~ exp (a) ((1/2) (a - a ^ 2)) #

# = e ^ a * (a / 2) * (1 - a) #