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Raghav Baheti Grade: 12
        

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

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

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


 

6 years ago
Saurabh Max
15 Points
										

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

6 years ago
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