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lim n->infinty (3^n+4^n)^1/n=
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
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
so answer is 4
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