showing limit of a sequence

[dusky nebula]

I having trouble showing this for all n members of the natural numbers. I need help

[shrewd lark]

what techniques were taught in class?

[dusky nebula]

@shrewd lark to be honest I dont know, I self study using abbot's textbook. In it he shows an N exists such that all for n>= N, a_n is in the epsilon-neighborhood of the claimed limit.

[shrewd lark]

oh ok that makes sense

so first things first: try to formulate this question in epsilon-delta language

$\forall \varepsilon > 0, \dots$

[errant prairie]

Dee3Cay

[gritty oasis]

$\sqrt[n]{a} = a^{\frac{1}{n}} = exp(\frac{1}{n} ln(a))$

[errant prairie]

Daddy_314

[gritty oasis]

Which means this converges to exp(0)

And if you're wondering why
$a^b = exp(b ln(a))$
It's because both have the same logarithm

[errant prairie]

Daddy_314

[gritty oasis]

Ln(a^b) = b ln(a)
Ln(exp(b ln(a)) = b ln(a)

[dusky nebula]

@gritty oasis what is the function exp(x), e^x ?. And do you have any tips for latex?

[gritty oasis]

Yes