When x→0 , lim (1 + 3ln2^x) ^ (1/x) =

lim (1 + 3ln2^x) ^ (1/x) = y
x→0

Ln from both sides ...

Answer in my book :
lny =lim ln(1+3ln2^x)/x (how?)
x→0
.....
answer is 8.

Let as x→0 lim (1 + 3ln2^x) ^ (1/x) = L

Then ln L = lim x-->0 ln ((1 + 3ln2^x) ^ (1/x)) = lim x-->0 ln ( 1+ 3 ln 2^x) * (1/x) (Since ln a^b= b lna)

So ln L= lim x-->0( ln (1 + 3 ln 2^x))/x = lim x-->0 (ln ( 1+ 3 x ln 2))/x

=lim x--> (((ln (1+ x 3ln2 ))/(x 3ln2)) 3ln2 (Multiplying and dividing by 3ln2.

as y-->0, lim (ln(1+y))/y =1

So ln L = 3ln 2 = ln 2^3 so L = 2^3 =8