The value oflimn→∞4√n5+2−3√n2+15√n4+2−2√n3+1is
0
limn→∞4√n5+2−3√n2+15√n4+2−2√n3+1=limn→∞n5/44√1+2n5−n2/33√1+1n2n4/55 ⎷1+2n4−n3/22√1+1n3=limn→∞n5/4n3/24√1+2n5−n2/3n3/23√1+1n2n4/5n3/25√1+2n4−n3/2n3/22√1+1n3On dividing the numerator and denominator by the highest power of n.i.e., n3/2=limn→∞1n1/44√1+2n5−1n5/63√1+1n2n7/105√1+2n4−2√1+1n3=0−00−1=0