Sequence's limit with Bernoulli's inequality

[dormar]

Hi there. I (and the rest of the class) have been hitting the wall with this one for days. Would love if someone here can help us with this.
The problem's is this:

Prove the following:
\(\displaystyle \lim_{n->\infty}{(2\sqrt[n]x-1)^n}=x^2\)
When x>1
Using the squeeze theorem and Bernoulli's inequality.

Now I've managed to prove x^2 is greater than the expression for large n, but no matter what permutation's of the expression I use Bernoulli's inequality with, I still don't get anything whose limit is x^2. Can anyone help?
Thank you so much.

BTW:
Bernoulli's inequality:

For \(\displaystyle x> -1\):
\(\displaystyle (1+x)^n \ge 1+nx\)

Also, I can't use advanced calculus methods, like l'Hospital's rule and the such, because we haven't studied them, we're just on sequence limits for now, with sequence arithmetic and the such.
Thanks again.

 

[galactus]

Try making a substitution. i.e. let t=1/n

I do not have time now to look at it in depth, but I will take a gander this afternoon.

Try working an LN and/or an e into your limit. Note that \(\displaystyle e^{2ln(x)}=x^{2}\)

If you can get it to this point, you have it made. Note that it looks similar to the famous e limits we learn about this time.

Such as \(\displaystyle \lim_{x\to 0}(1+x)^{\frac{1}{x}}=e\)

or \(\displaystyle \lim_{x\to {\infty}}(1+\frac{1}{x})^{x}=e\)

 

[dormar]

Thanks for the reply!
Unfortunately, we're not supposed to use limits of numbers approaching 0 (as you have suggested, I assume - calling t=1/n and then taking the limit as it approaches 0). It might help, but we're supposed to only use sequences using a natural number (n) approaching infinity.
As for the e^2lnx, I've been trying to get to that expression, without success. Also, I'm pretty sure it can't be done - not with bernoulli anyway, since e^2lnx=x^2 which is the series approaching x^2 that is greater then the mentioned series, and we're looking for the series that's always lesser than the given series.
With many thanks, again...
Dor.