How do i proof [math]\lim_n a_n = a \Rightarrow \lim_n \sqrt[m]{a_n}

>>8078298
Can you answer my question? I seriously can't find a proof. I thought this is the math related board.

>>8078265
The root function is continuous.
For the proof of that you should look at exp(1/m * ln(a)) which is the m-th root of a. It's obviously a composition of continuous functions and by that also continuous.

>>8078323
Thanks.

>>8078323

Here you're proving continuity of the m-th root by assuming continuity of exp and log. If it's "obvious" that exp and log are continuous, why is it not "obvious" that the mth root is continuous?

>>8078372
It is
The proof for the continuity of exp and log is trivial

>>8078389

But then, by your proof above, the continuity of x^(1/m) is trivial.

So your entire proof has been reverted to "it's trivial".

Proving that exp(x) is continuous is far less trivial than the proof you have given above.

>>8078372
A^(x+r)=A^x * A^r
Just find the limit for aproaching 0 and voila exp function is continuous

>>8078399
it is trivial once you are no longer in HS

>>8078404
>>8078407

Yes geniuses, it does reduce to proving that lim x-->0 of exp(x) = 1.

Now prove this without using continuity of exp(x).

>>8078407

Big talk. Now prove it, without finding the proof online.

Clown.

>>8078420
WTF? Are you a fucking retard? How old are you?
A^(x+0)=A^x * A^0

>>8078577
>>8078428
>>8078420
>>8078407
>>8078404
>>8078399
>>8078389
>>8078372
>>8078372
OP here, no more problem, i got the proof :)