# limit to log

• Oct 24th 2007, 07:05 AM
BrainMan
limit to log
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) = x

b) [(1 + x^(1/n))/2]^(n) = 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.

• Oct 24th 2007, 07:59 AM
CaptainBlack
Quote:

Originally Posted by 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) = x

b) [(1 + x^(1/n))/2]^(n) = 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.

What are the "=" signs doing here? Is x changing with n?

RonL
• Oct 24th 2007, 09:04 AM
ThePerfectHacker
Quote:

Originally Posted by 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) = x

b) [(1 + x^(1/n))/2]^(n) = 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.

You can write,
$(1+x^{1/n})/2 = 1+a_n/n$ where $a_n = n(x^{1/n}-1)/2$. So now you need to compute $(1+a_n/n)^{1/n} \to e^a$ where $a=\lim a_n$. Thus, it remains to find $n(x^{1/n}-1)\to \log x$
• Oct 24th 2007, 09:42 AM
BrainMan
Quote:

Originally Posted by CaptainBlack
What are the "=" signs doing here? Is x changing with n?

RonL

That's what the limits equal. Sorry about the confusion.

I need the limit of [(1 + x^(1/n))/2]^(2n), which equals x, and the limit of [(1 + x^(1/n))/2]^(n), which equals the square root of x.

I just don't know how to show them with algebra.

PerfectHacker, thanks for the response, but I don't really see what you're doing. Could you elaborate? Basically, how does it simplify to x?
• Oct 24th 2007, 11:44 AM
BrainMan
Ignore the last post. Looking at it more closely, I see what you did. However, how do you know the last limit goes to ln(x)? Could you just explain that?
• Oct 24th 2007, 04:51 PM
ThePerfectHacker
Quote:

Originally Posted by BrainMan
Ignore the last post. Looking at it more closely, I see what you did. However, how do you know the last limit goes to ln(x)? Could you just explain that?

Yes. This limit is all that remains.

$x^{1/n} - 1 = [1+(x-1)]^{1/n} - 1 = \frac{1}{n}(x-1) + \frac{1}{n} \left( \frac{1}{n} - 1 \right)\frac{(x-1)^2}{2!}+$ $\frac{1}{n} \left(\frac{1}{n} - 1 \right) \left( \frac{1}{n} - 2\right) \frac{(x-1)^3}{3!}+...$

So $n(x^{1/n}-1)$ is,
$(x-1) + \left( \frac{1}{n} - 1 \right)\frac{(x-1)^2}{2!}+$ $\left(\frac{1}{n} - 1 \right) \left( \frac{1}{n} - 2\right) \frac{(x-1)^3}{3!}+...$
As we take the limit we end up with,
$\frac{(x-1)}{1} - \frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-... = \log x$

This is Mine 74:):)th Post!!!