Koja je granica ove funkcije s h približava 0? (H) / (sqrt (4 + h) -2)

#Lt_ (h-> o) (h) / (sqrt (4 + h) -2) #

# = Lt_ (h-> o) (h (sqrt (4 + H) + 2)) / ((sqrt (4 + h) -2) (sqrt (4 + H) + 2) *

# = Lt_ (h-> o) (h (sqrt (4 + H) + 2)) / (4 + H-4) #

# = Lt_ (h-> o) (poništi (sqrt (4 + h) +2)) / cancelh "as" h! = 0 #

# = (Sqrt (4 + 0) + 2) = 2 + 2 = 4 #

Odgovor:

# 4#.

Obrazloženje:

Sjetite se da, #lim_ (h do 0) (f (a + h) -f (a)) / h = f '(a) ………… (ast) #.

Neka, #f (x) = sqrtx, "tako da," f '(x) = 1 / (2sqrtx) #.

#:. f '(4) = 1 / (2sqrt4) = 1/4 #.

Ali, # f '(4) = lim_ (h do 0) (sqrt (4 + h) -sqrt4) / h ………… jer, (ast) #.

#:. lim_ (h do 0) (sqrt (4 + h) -sqrt4) / h = 1/4 #.

#:. "Reqd. Lim." = 1 / (1/4) = 4 #.

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