Lim_ (x-> 0) (sqrt (1 + x ^ 2) -sqrt (1 + x)) / (sqrt (1 + x ^ 3) -sqrt (1 + x)) =?

Lim_ (x-> 0) (sqrt (1 + x ^ 2) -sqrt (1 + x)) / (sqrt (1 + x ^ 3) -sqrt (1 + x)) =?
Anonim

Antwoord:

#lim_ (x-> 0) (sqrt (1 + x ^ 2) -sqrt (1 + x)) / (sqrt (1 + x ^ 3) -sqrt (1 + x)) = 1 #

Uitleg:

Met de regel van L'Hopital weten we dat #lim_ (x-> a) (f (x)) / (g (x)) => (f (a)) / (g (a)) #

#f (x) = sqrt (1 + x ^ 2) -sqrt (1 + x) #

# = (1 + x ^ 2) ^ (1/2) - (1 + x) ^ (1/2) #

#f '(x) = x (1 + x ^ 2) ^ (- 02/01) - (1 + x) ^ (- 02/01) / 2 #

#G (x) = sqrt (1 + x ^ 3) -sqrt (1 + x) #

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

#G '(x) = (3x ^ 2 (1 + x ^ 3) ^ (- 02/01)) / 2- (1 + x) ^ (- 02/01) / 2 #

#lim_ (x-> 0) (sqrt (1 + x ^ 2) -sqrt (1 + x)) / (sqrt (1 + x ^ 3) -sqrt (1 + x)) => (0 (1+ 0 ^ 2) ^ (- 02/01) - (1 + 0) ^ (- 02/01) / 2) / ((3 (0) ^ 2 (1 + 0 ^ 3) ^ (- 02/01)) / 2- (1 + 0) ^ (- 1/2) / 2) #

#=(-(1+0)^(-1/2)/2)/(-(1+0)^(-1/2)/2)#

# = Annuleren (- (1 + 0) ^ (- 02/01) / 2) / annuleren (- (1 + 0) ^ (- 02/01) / 2) = 1 #