Hvordan finner du lim_ (xtooo) logg (4 + 5x) - logg (x-1)?

Hvordan finner du lim_ (xtooo) logg (4 + 5x) - logg (x-1)?
Anonim

Svar:

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

Forklaring:

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

Bruke kjederegel:

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

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

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

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