Răspuns:
c) Aplicăm L'H și
obținem -2/6 = -1/3.
f) lim (6*x^2-3*x+11)'/(2*x+6)'
lim (6*x^2-3*x+11)'/(2*x+6)' x–> -∞
=(12*x-3)/2=12/0= -∞.
c) lim (x+√x)/(3*x+2*√x+1)
) lim (x+√x)/(3*x+2*√x+1) x–> ∞
= lim (1+1/(2*√x))/(3+1/√x)
x–>∞
= lim (2√x+1)/(2*√x)/(3*√x+1)/(√x)
x–>∞
= lim (2√x+1)/(6√x+2)
lim (2√x+1)/(6√x+2) x–>∞
= lim (1/√x)/(3/√x)
lim (1/√x)/(3/√x) x–>∞
= 1/3.
