Question

limita xtinde5 x-5 pe radical 2x-1-3 (3 inafara radicalului)
limita xtinde0 radical x+9-3(3 inafara radicalului) pe x2+2x
Helppppp please!!!

Answer 1

prima  limta..  Amplifici  fractia cu ( √(2x-1)+3
(x-5)*(√(2x-1)+3)/(2x-1-9)=(x-5)*(√(2x-1)+1)/(2x-1-3)=(x-5)(√(2x-1)+1)/2(x-5)=(√(2x-1)+3)/2
treci  la  limita
lim (√2x-1)+1)/2=(√9+1)/2=10/2=5  x→5
b)  amplifici  fractia  cu √(x+9)+3
(√(x+9-3)*(√x+9)+3)/x(x+2)*(√x+9)+3)=(x+9-9)/x*(x+2)*√(x+9)+3=x/x*(x+2)*(√(x+9)+3)=1/(x+2)*(√(x+9)+3
lim 1/(x+2)*(√(x+9)+3)=1/2*6=1/12 cand  x→0