คุณหาข้อ จำกัด ได้อย่างไร (X-> 0) ขอบคุณ

ตอบ:

#sqrt (6) #

คำอธิบาย:

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

# = 1 + x * ln (a) + (x * ln (a)) ^ 2/2 + (x * ln (a)) ^ 3/6 + … #

# => 3 ^ x + 2 ^ x = 2 + x * (ln (3) + ln (2)) + x ^ 2 * (ln (3) ^ 2 + ln (2) ^ 2) / 2 + x ^ 3 * (ln (3) ^ 3 + ln (2) ^ 3) / 6 + … #

# = 2 + x * ln (6) + x ^ 2 * (… #

# => (3 ^ x) ^ 2 + (2 ^ x) ^ 2 = 3 ^ (2x) + 2 ^ (2x) #

# = 2 + 2 * x * ln (6) + 4 * x ^ 2 * (ln (2) ^ 2 + ln (3) ^ 2) / 2 + 8 * x ^ 3 * (ln (3) ^ 3 + ln (2) ^ 3) / 6 + … #

# => (3 ^ (2x) + 2 ^ (2x)) / (3 ^ x + 2 ^ x) = "#

# 1 + (x * ln (6) + 3 * x ^ 2 * …) / (2 + x * ln (6) + x ^ 2 * …) #

# ~~ 1+ (x * ln (6)) / 2 "(สำหรับ x" -> "0)" #

# "ยกกำลัง 1 / x ให้ผลตอบแทน:" #

# (1+ (x * LN (6)) / 2) ^ ((2 / (x * LN (6))) * (LN (6) / 2)) #

# = e ^ (ln (6) / 2) #

# = sqrt (6) #