How would you logically derive lim(x->infinity) x^(-x)?

Question

Is it 0 or undetermined (and you would have to use LaHospital's (sp?))? thanks

Actually, the problem was not x^(-x) but x^(1/x)...i tihnk it still works.

I see, mathdumb; your notation was unclear. thanks i got it though

And when you have indetermine as your answer like inf^inf or inf/inf; you use L'Hospital's rule. math dumbs got it i think.

Answer 1

Write x^(-x) as 1/(x^x) and use L'Hospitals' (note spelling ☺) rule to show it's 0.


Doug

Answer 2

infinity and detrimental infinity are not from now on numbers. We use the be conscious "infinity" after we keep in mind that as x receives very on the fringe of 0, the expression grows with out certain yet detrimental. try some values of x that are on the fringe of 0. try 0.0001, then 0.00001, etc. The expression must be increasing yet detrimental I propose utilising a calculator. playstation : a million/0 isn't a defined determination!!!

Answer 3

Given:
x^1/x as x → ∞ then;
The base becomes ∞ and the power becomes 0.
→ ∞º But if the base is not zero then anything to power zero is 1.
→ ∞º = 1

Answer 4

x^(-x) = 1/(x^x)

since anything raised to 0 will always get you 1,

x has no limitations

Range : 1 <= y < 0

Answer 5

You dont need lHopital. 1/x^x is 1 / something that get infinit big. Imagine 1 dolar for the whole inhabitants. How much would you get?

Ana

Answer 6

If you write it as 1/x^x, the L'Hopital's rule isn't needed (and doesn't apply). The limit as x->+infty will be 0.

Answer 7

it is x^(1/x)
take log we get 1/x(logx) logx/x
as both tend to infinite use the rule

it is 1/x/1 or 1/x
it tends to 0
so value tends to 1 as log tends to 0

Answer 8

i think results given are fault; i think u should think this:
x^(1/x)=1/x.exp(ln(x)) which tends to infinity on infinity, which is underterminate