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 e^log(3/x²+4/x)^x

L = limit  e^xlog(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