Berapa batas fungsi ini ketika h mendekati 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) (cancelh (sqrt (4 + h) +2)) / cancelh "as" h! = 0 #

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

Menjawab:

# 4#.

Penjelasan:

Ingat itu, #lim_ (h ke 0) (f (a + h) -f (a)) / h = f '(a) ………… (ast) #.

Membiarkan, #f (x) = sqrtx, "jadi itu," f '(x) = 1 / (2sqrtx) #.

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

Tapi, # f '(4) = lim_ (h to 0) (sqrt (4 + h) -sqrt4) / h ………… karena, (ast) #.

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

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

Nikmati Matematika.!