- Thread starter Ovan
- Start date

- Joined
- Jun 18, 2007

- Messages
- 19,933

Please follow the rules of posting in this forum, as enunciated at:

Please share your

Hint: start with:

\(\displaystyle \left [\frac{\sqrt[n]{3} + \sqrt[n]{4}}{2} \right ] ^n\)

= \(\displaystyle (4)*\left [\frac{\sqrt[n]{\frac{3}{4}} + 1}{2} \right ] ^n\)

Continue......

I really don't know how to continue.. can u show me some more steps, please?View attachment 15967

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share yourwork/thoughtsabout this assignment

Hint: start with:

\(\displaystyle \left [\frac{\sqrt[n]{3} + \sqrt[n]{4}}{2} \right ] ^n\)

= \(\displaystyle (4)*\left [\frac{\sqrt[n]{\frac{3}{4}} + 1}{2} \right ] ^n\)

Continue......

- Joined
- Jun 18, 2007

- Messages
- 19,933

Before I show more steps, please tell me:I really don't know how to continue.. can u show me some more steps, please?

What math class are you taking now (or have taken recently) - e.g.- precalculus, advanced algebra, calculus-I, etc.?

Have you studied Taylor's expansion formula?

Have you studied formula for Binomial expansion?

I have studied both but I don't know how to use it in this example.Before I show more steps, please tell me:

What math class are you taking now (or have taken recently) - e.g.- precalculus, advanced algebra, calculus-I, etc.?

Have you studied Taylor's expansion formula?

Have you studied formula for Binomial expansion?

- Joined
- Jun 18, 2007

- Messages
- 19,933

Please expand using Binomial expansion theorem (first few terms):I have studied both but I don't know how to use it in this example.

(1 + x)

- Joined
- Nov 24, 2012

- Messages
- 2,351

\(\displaystyle L=\lim_{n\to\infty}\left(\left(\frac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)^n\right)\)

Taking the natural log of both sides, we eventually get:

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(\dfrac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)}{\dfrac{1}{n}}\right)\)

Now you have the indeterminate for 0/0 and may apply L'Hôpital's Rule.

- Joined
- Jun 18, 2007

- Messages
- 19,933

Again assumed things that I should not have assumed. Somehow I thought L'Hospital was not allowed (we had spate of those sadistic problems in near-past!)

\(\displaystyle L=\lim_{n\to\infty}\left(\left(\frac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)^n\right)\)

Taking the natural log of both sides, we eventually get:

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(\dfrac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)}{\dfrac{1}{n}}\right)\)

Now you have the indeterminate for 0/0 and may apply L'Hôpital's Rule.

(1 + x)Please expand using Binomial expansion theorem (first few terms):

(1 + x)^{n}= ?

\(\displaystyle L=\lim_{n\to\infty}\left(\left(\frac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)^n\right)\)

Taking the natural log of both sides, we eventually get:

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(\dfrac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)}{\dfrac{1}{n}}\right)\)

Now you have the indeterminate for 0/0 and may apply L'Hôpital's Rule.

- Joined
- Nov 24, 2012

- Messages
- 2,351

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\dfrac{1}{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}\left(-n^{-2}\left((3)^{\frac{1}{n}}\ln(3)+(4)^{\frac{1}{n}}\ln(4)\right)\right)}{-n^{-2}}\right)\)

Hence:

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\dfrac{(3)^{\frac{1}{n}}\ln(3)+(4)^{\frac{1}{n}}\ln(4)}{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}\right)=\ln\left(2\sqrt{3}\right)\)

And so we conclude:

\(\displaystyle L=2\sqrt{3}\)

- Joined
- Jun 18, 2007

- Messages
- 19,933

First you should observe that:

\(\displaystyle ln \left[ \dfrac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right] = ln[(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}] - ln(2)\)

and

\(\displaystyle \frac{d}{dn}\left[ \frac{1}{n}\right ]\) = - \(\displaystyle \ \frac{1}{n^2}\)

now continue......

- Joined
- Jun 18, 2007

- Messages
- 19,933

It will help you to build "your character" (according to my UG calculus professor).

Or another idea is performing the change of variable x=1/n to MarkFL's suggestion which makes applying L'Hôpital's rule easy. I ended up with...\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(\dfrac{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}{2}\right)}{\dfrac{1}{n}}\right)\)

Now you have the indeterminate for 0/0 and may apply L'Hôpital's Rule.

\(\displaystyle \ln(L) =\lim_{x\to0}\left( \frac{3^x\cdot \ln\left(3\right)+2\mathinner{\times} 4^x\cdot \ln\left(2\right)}{3^x+4^x} \right) \)

and from there it's simple!

- Joined
- Nov 24, 2012

- Messages
- 2,351

It appeared to me that the OP was actually already at the doorstep in the first attached image, having gotten to this point:

It will help you to build "your character" (according to my UG calculus professor).

\(\displaystyle \ln(L)=\lim_{n\to\infty}\left(\dfrac{(3)^{\frac{1}{n}}\ln(3)+(4)^{\frac{1}{n}}\ln(4)}{(3)^{\frac{1}{n}}+(4)^{\frac{1}{n}}}\right)\)

Now, we only have to recognize this is a determinate form to complete the problem.

\(\displaystyle \ln(L)=\frac{\ln(3)+\ln(4)}{2}=\frac{1}{2}\ln(12)=\ln(\sqrt{12})=\ln(2\sqrt{3})\)

And the conclusion I posted above then follows.