vikas askiitian expert
Last Activity: 13 Years ago
L = (3n2+4n)1/n lim n=>infinity
put n = 1/x
L = limit (3/x2+4/x)x
x=>0
can be written as elog(3/x2+4/x)x
L = limit exlog(3/x2+4/x) x=>0
L = limit e[log(3/x2+4/x) / 1/x] x=>0
this limit is (infinity/infinity) form so we can use L holpital rule
differentiate numerator & denominator with respect to x & put the limit
L = e0 = 1