limits: [(1+x^(1/n))/2]^(2n) for x>1, as n->infty

[BrainMan]

Could someone show me how to solve the following limits algebraically:

For x > 1, lim as n approaches infinity of

a) [(1 + x^(1/n))/2]^(2n), which equals x
b) [(1 + x^(1/n))/2]^(n), which equals sqrt(x)

They seem very similar to the sequence (1 + (a/n))^n whose limit is e^a, but I don't see how to manipulate it accordingly.

If you could show me how to solve these, I would really appreciate it.

Thanks in advance.

 

[galactus]

Here's a rather comlicated method I worked out. Hope it helps. I went ahead and broke the unwritten rule by using L'Hopital.

\(\displaystyle \L\\\lim_{n\to\infty}\left(\frac{1}{2}+\frac{1}{2}x^{\frac{1}{n}}\right)^{2n}\)

\(\displaystyle \L\\\lim_{n\to\infty}e^{\left(2nln\left(\frac{1}{2}+\frac{1}{2}x^{\frac{1}{n}}\right)\right)}\)

Now, use L'Hopital:

\(\displaystyle \L\\e^{\left(2\lim_{n\to\infty}\frac{x^{\frac{1}{n}}ln(x)}{1+x^{\frac{1}{n}}\right)\)

\(\displaystyle \L\\e^{\left(2ln(x)\frac{\lim_{n\to\infty}x^{\frac{1}{n}}}{1+\lim_{n\to\infty}x^{\frac{1}{n}}}\right)}\)

Now, notice the limit in the denominator?. It tends to 1 as n increases without bound. Therefore, the 2's cancel:

\(\displaystyle \L\\e^{\left(ln(x)\lim_{n\to\infty}x^{\frac{1}{n}}\right)\)

\(\displaystyle \L\\e^{ln(x)(1)}\)

\(\displaystyle \H\\=\fbox{x}\)


For the second one, notice it follows from the first.

The limit of the first is x, so:

\(\displaystyle \L\\\left(\frac{1}{2}+\frac{1}{2}x^{\frac{1}{n}}\right)^{2n}=\left(\underbrace{\left(\frac{1}{2}+\frac{1}{2}x^{\frac{1}{2}}\right)^{2n}}_{\text{this is x}}\right)^{\frac{1}{2}}=\left(\frac{1}{2}+\frac{1}{2}x^{\frac{1}{2}}\right)^{n}=\sqrt{x}\)

 

[BrainMan]

Thanks a lot. I really appreciate your help.