Răspuns
∞
Explicație pas cu pas:
lim x->π/2 x<π/2 dim (cos x),=+0⇒lim x->π/2 x<π/2din (1/cosx)=1/+0=∞ si
lim x->π/2 x<π/2din (1/cosx)²=∞²=∞
lim x->π/2 x>π/2 dim (cos x),=-0⇒lim x->π/2 x>π/2din (1/cosx)=1/-0=-∞
si
lim x->π/2 x<π/2din (1/cosx)²=(-∞)²=∞
deci limita exista si este ∞
analog avem( (2/(+0))²=∞²=∞
si, respectiv
(2/(-0))²=(-∞)²=∞
deci limita exista si este ∞
