Question

Question #0f1e0

Answer 1

`-1`

Explanation:

`lim_(nrarroo)-n^(1/n)`

The `1/n` term will approach `0` as `nrarroo`. Few examples:

`1/2=0.5`
`1/3=0.333...`
`1/5=0.2`
`1/10=0.1`
`1/100 = 0.01`

As you can see, as the denominator gets bigger, it approaches 0.

This means that our question becomes `lim_(nrarroo)-n^0`

Since any number to the power of `0` is `1`, all that is left is the `-` sign and the `1`.

`lim_(nrarroo)-n^0`

`=lim_(nrarroo)-1`

`=-1`

Answer 2

-1, using `ln` and L'opitals rule

Explanation:

First, lets try plugging in the limit: `-n^(1/n)-> -oo^0`.
This produces an indeterminate form, in other words we can't draw any immediate conclusions from this; we have to do more work.

Whenever we have one function to the power of another (in this case `n` to the `1/n`), it usually helps to let the set the limit equal to some variable `L`, then take then `ln` of both sides, as shown below. We'll leave out the - sign as well for now, just remembering to change the sign of the final answer:

`L=lim_(n->oo)n^(1/n)`
`ln(L)=lim_(n->oo)ln(n^(1/n))` (Side note: we can do this because the ln function is continuous for the region we're interested in)
But by log laws, we can move the index down:
`ln(L)=lim_(n->oo)(1/n)ln(n)`
`ln(L)=lim_(n->oo)ln(n)/n`

Now if we plug in `oo` to the left, we get the form `oo/oo`, and because the numerator and denominator are both continuous functions we get to use L'opitals rule:

`ln(L)=lim_(n->oo)(1/n)/1`
`ln(L)=lim_(n->oo)1/n`
NOW, the right hand side is for sure 0:
`ln(L) = 0`
And we raise both sides to `e`:
`L=e^0=1`

But remember that there was a - sign we left out for convenience, so we put that back in and get our final answer of -1

P.S. I'm assuming you means `-(n^(1/n))` not `(-n)^(1/n)`, because otherwise we wouldn't get any answer at all.

If any of the above things such as L'opitals rule need further explanation, please ask.