How do you solve #lim_(n->oo) (3^n + 5^n + 7^n)^(1/n) = #?
1 Answer
# lim_(n rarr oo) (3^n+5^n+7^n)^(1/n) = 7 #
Explanation:
We have:
# L = lim_(n rarr oo) (3^n+5^n+7^n)^(1/n) #
As the logarithmic function is monotonic, then we can take the log of the limit, to get:
# ln L = ln lim_(n rarr oo) (3^n+5^n+7^n)^(1/n) #
# " " = lim_(n rarr oo) ln {(3^n+5^n+7^n)^(1/n)} #
# " " = lim_(n rarr oo) 1/n \ ln (3^n+5^n+7^n) #
# " " = lim_(n rarr oo) (ln (3^n+5^n+7^n))/n #
This is now of an indeterminate form
# ln L = lim_(n rarr oo) (d/(dn) \ ln (3^n+5^n+7^n))/(d/(dn) n) #
# " " = lim_(n rarr oo) (((ln3)3^x + (ln5)5^x + (ln7)7^x)/(3^n+5^n+7^n))/1 #
# " " = lim_(n rarr oo) ((ln3)3^n + (ln5)5^n + (ln7)7^n)/(3^n+5^n+7^n)#
Now we can simplify the limit by multiplying the numerator and denominator by
# ln L = lim_(n rarr oo) ((ln3)3^n + (ln5)5^n + (ln7)7^n)/(3^n+5^n+7^n) * (1/7^n)/(1/7^n) #
# " " = lim_(n rarr oo) ((ln3)3^n/7^n + (ln5)5^n/7^n + (ln7)7^n/7^n)/(3^n/7^n+5^n/7^n+7^n/7^n) #
# " " = lim_(n rarr oo) ((ln3)(3/7)^n + (ln5)(5/7)^n + (ln7)(7/7)^n)/((3/7)^n+(5/7)^n+(7/7)^n) #
# " " = lim_(n rarr oo) ((ln3)(3/7)^n + (ln5)(5/7)^n + ln7)/((3/7)^n+(5/7)^n+1) #
Now, we note that as both
# (3/7)^n # and#(5/7)^n rarr 0# as#n rarr oo#
Thus we can evaluate our limit to get:
# ln L = ((ln3)(0) + (ln5)(0) + ln7)/(0+0+1) #
# " " = ln 7 #
From which we conclude that:
# L = 7 #