Kako najdete lim_ (xtooo) log (4 + 5x) - log (x-1)?

Kako najdete lim_ (xtooo) log (4 + 5x) - log (x-1)?
Anonim

Odgovor:

#lim_ (xtooo) log (4 + 5x) - log (x-1) = dnevnik (5) #

Pojasnilo:

#lim_ (xtooo) log (4 + 5x) - log (x-1) = lim_ (xtooo) log ((4 + 5x) / (x-1)) #

Z uporabo verižnega pravila:

#lim_ (xtooo) log ((4 + 5x) / (x-1)) = lim_ (utoa) log (lim_ (xtooo) (4 + 5x) / (x-1)) #

#lim_ (xtooo) (ax + b) / (cx + d) = a / c #

#lim_ (xtooo) (5x + 4) / (x-1) = 5 #

#lim_ (uto5) dnevnik (u) = log5 #