ทำไม lim_ (x-> oo) (sqrt (4x ^ 2 + x-1) -sqrt (x ^ 2-7x + 3)) = lim_ (x-> oo) (3x ^ 2 + 8x-4) / ( 2x + ... + x + ... ) = OO?

ตอบ:

# "ดูคำอธิบาย" #

คำอธิบาย:

# "คูณด้วย" #

# 1 = (sqrt (4 x ^ 2 + x - 1) + sqrt (x ^ 2 - 7 x + 3)) / (sqrt (4 x ^ 2 + x - 1) + sqrt (x ^ 2 - 7 x + 3)) #

# "จากนั้นคุณจะได้รับ" #

#lim_ {x-> oo} (3 x ^ 2 + 8 x - 4) / (sqrt (4 x ^ 2 + x - 1) + sqrt (x ^ 2 - 7 x + 3)) #

# "(เพราะ" (a-b) (a + b) = a ^ 2-b ^ 2 ")" #

# = lim_ {x-> oo} (3 x ^ 2 + 8 x - 4) / (sqrt (4 x ^ 2 (1 + 1 / (4x) - 1 / (4x ^ 2))) + sqrt (x ^ 2 (1 - 7 / x + 3 / x ^ 2)) #

# = lim {x-> oo} (3 x ^ 2 + 8 x - 4) / (2x sqrt (1 + 0 - 0) + x sqrt (1 - 0 + 0)) #

# "(เพราะ" lim_ {x-> oo} 1 / x = 0 ")" #

# = lim {x-> oo} (3 x ^ 2 + 8 x - 4) / (3 x) #

# = lim {x-> oo} (x + (8/3) - (4/3) / x) #

# = oo + 8/3 - 0 #

# = OO #