Pt a,b,c se aplica formula lim sinu/u=1 pt u→0
a) lim sin5x/lim5*sin5x/5x=5lim sin5x/5x=5*1=5 x→0
b)lim sin2x/7x= lim 2/7*7/2*sin2x/7x=2/7*lim sin2x/2x=2/7*1=2/7
c)lim (sin3x+sin5x)/x=lim[sin3x/x+sin5x/x]
lim sin3x/x=lim 3*sin 3x/3x=3lim sin3x/3x=3*1=3
lim sin5x/x=lim 5*sin5x/5x=5*limsin5x/5x=5
L=3+5=8
b)sus Trecand la limita ajungi la 0/0.aplici teorema lui LHospital
L=lim[(2-√(x-3)] `/(x²)`= -1/2√(x-3)*1/2x= -1/√(7-3)*1/2*7= -1/√4*1/14=-1/2*1/14=-1/28
d)Consideri numarul ca o fractie cu numitorul 1.Amplifici fractia cu x-√(x²+1)
lim(x²-(x²+1))/[x+√(x²+1)=lim1/[x-√(x²+1)]=lim 1/[x-lxl*√(1+1/x²)=lim1/[x+x√(1+1/x²)=1/-∞=0 1/x²→0
