Question

How do you solve `lim_(n->oo) (3^n + 5^n + 7^n)^(1/n) =`?

Answer 1

`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 `oo//oo`, so we can apply L'Hôpital's rule to get

`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 `1/7^n`, to get:

`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| lt 1` and `|5/7| lt 1` then both entities:

`(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`