Question

How do you determine the limit of `[(n+5)/(n-1)]^n` as n approaches `oo`?

Answer 1

`lim_(n->oo) ((n+5)/(n-1))^n = e^6`

Explanation:

Write the expression as:

`[(n+5)/(n-1)]^n = [(n-1+6)/(n-1)]^n = (1+ 6/(n-1))^n`

Now take the logarithm:

`ln ((n+5)/(n-1))^n = n ln (1+ 6/(n-1))`

and divide and multiply by `6/(n-1)`:

`ln ((n+5)/(n-1))^n = 6(n/(n-1))( (ln (1+ 6/(n-1)))/(6/(n-1)))`

Now consider the succession: `b_n = 6/(n-1)`. Clearly:

`lim_(n->oo) b_n = 0`

so that :

`lim_(n->oo) (ln (1+ 6/(n-1)))/(6/(n-1)) = lim_(x->0) ln(1+x)/x = 1`

As:

`lim_(n->oo) n/(n-1) = 1`

as well, we have then:

`lim_(n->oo) ln ((n+5)/(n-1))^n = 6`

Now, using the continuity of the exponential function:

`lim_(n->oo) ((n+5)/(n-1))^n = lim_(n->oo) e^(ln ((n+5)/(n-1))^n) = e^(lim_(n->oo) ln ((n+5)/(n-1))^n) = e^6`