# Thread: limit to log

1. ## 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.

Thanks in advance.

2. 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.

Thanks in advance.
What are the "=" signs doing here? Is x changing with n?

RonL

3. 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.

Thanks in advance.
You can write,
$\displaystyle (1+x^{1/n})/2 = 1+a_n/n$ where $\displaystyle a_n = n(x^{1/n}-1)/2$. So now you need to compute $\displaystyle (1+a_n/n)^{1/n} \to e^a$ where $\displaystyle a=\lim a_n$. Thus, it remains to find $\displaystyle n(x^{1/n}-1)\to \log x$

4. 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?

5. 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?

6. 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.

$\displaystyle 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!}+$$\displaystyle \frac{1}{n} \left(\frac{1}{n} - 1 \right) \left( \frac{1}{n} - 2\right) \frac{(x-1)^3}{3!}+... So \displaystyle n(x^{1/n}-1) is, \displaystyle (x-1) + \left( \frac{1}{n} - 1 \right)\frac{(x-1)^2}{2!}+$$\displaystyle \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,
$\displaystyle \frac{(x-1)}{1} - \frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-... = \log x$

This is Mine 74th Post!!!