Question

How do you determine the limit of `(7(e^(5n))+5n)^(1/n)` as n approaches infinity?

Answer 1

`lim_(nrarroo)(7e^(5n)+5n)^(1/n)=e^5`

Explanation:

Let

`L=lim_(nrarroo)(7e^(5n)+5n)^(1/n)`

Taking the natural logarithm of both sides gives

`ln(L)=ln(lim_(nrarroo)(7e^(5n)+5n)^(1/n))`

Since `ln(x)` is a continuous function this can become

`ln(L)=lim_(nrarroo)ln((7e^(5n)+5n)^(1/n))`

Rewriting using logarithm rules,

`ln(L)=lim_(nrarroo)1/nln(7e^(5n)+5n)`

`ln(L)=lim_(nrarroo)ln(7e^(5n)+5n)/n`

The limit is in the indeterminate form `oo/oo` so L'Hospital's rule applies, yielding

`ln(L)=lim_(nrarroo)(d/(dn)ln(7e^(5n)+5n))/(d/(dn)n)`

`ln(L)=lim_(nrarroo)((35e^(5n)+5)/(7e^(5n)+5n))/1`

`ln(L)=lim_(nrarroo)(35e^(5n)+5)/(7e^(5n)+5n)`

We see that `e^(5n)` is the overpowering term in the limit here, so the limit as `nrarroo` will be `(35e^(5n))/(7e^(5n))=5`. If you are dissatisfied with this explanation, reapplying L'Hospital's rule for two more iterations will get you to the same answer of `5`.

`ln(L)=5`

Rearranging to solve for the limit shows

`L=e^5`