Hoe vind je lim_ (xtooo) log (4 + 5x) - log (x-1)?

Hoe vind je lim_ (xtooo) log (4 + 5x) - log (x-1)?
Anonim

Antwoord:

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

Uitleg:

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

Chain rule gebruiken:

#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) log (u) = log5 #