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# lim n->infinty (3^n+4^n)^1/n=

10 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

10 years ago

U can do this

take 4^n common n out of bracket

due to^1/n it will become 4

we will have lim [1+(3/4)^n]^1/n*4

now 3/4 is a fraction hence its power infinity will be zero,u can check this from graph

and whole bracket raised to 1/infinity will be 1 as it is as good as raised to zero