what is limit of (1+ax)^1/xhen x → ∞ ????????????
0
use e^fx form
take log u getlog y=1/x log(1+ax)
using l hospital u get
logy=-1/x2 X a/1+axsoputting x->∞
we getlogy=0y=1
but answer is e^a
(1+ax)^1/x
= {(1+ax)^1/(ax)}^a
= e^a
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