Kakšna je meja te funkcije, ko se h približuje 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) (prekliči (sqrt (4 + h) +2)) / cancelh "as" h! = 0 #

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

Odgovor:

# 4#.

Pojasnilo:

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

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

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

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

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

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

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