Question

求解大一高数极限题

求具体过程

Answer 1

(1)

lim(x->铅睁+∞) [(x^2+1)/(x^3+x)]=1 and |3+cosx|≤1
=>
lim(x->+∞) [(x^2+1)/(x^3+x)] [3+(cosx)] =0

(2)
lim(x->∞禅裤) [x^2 -5(cosx)]/[3x^2+6(sinx)]
=lim(x->贺激简∞) (x^2)/(3x^2)
=1/3