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

Question

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

vikas askiitian expert · Answer

L = (3n 2 +4n) 1/n lim n=>infinity put n = 1/x L = limit (3/x 2 +4/x) x x=>0 can be written as e log (3/x 2 +4/x) x L = limit e xlog (3/x 2 +4/x) x=>0 L = limit e [log (3/x 2 +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 = e 0 = 1

Saurabh Max · Answer

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

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

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