Limit problem

Ovan

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Subhotosh Khan

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

READ BEFORE POSTING

Please share your work/thoughts about 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......
 

Ovan

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View attachment 15967
Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about 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......
I really don't know how to continue.. can u show me some more steps, please?
 

Subhotosh Khan

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

Ovan

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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?
I have studied both but I don't know how to use it in this example.
 

Subhotosh Khan

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

(1 + x)n = ?
 

MarkFL

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I would state:

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

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. :)
 

Subhotosh Khan

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I would state:

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

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. :)
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!)
 

Ovan

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

(1 + x)n = ?
(1 + x)n=
I would state:

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

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. :)
82203587_701875533674114_4866894551408705536_n.jpg82845721_3293799597303793_6716760240223682560_n.jpg
 

MarkFL

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I would continue from my previous post as follows:

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

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:

\(\displaystyle L=2\sqrt{3}\)
 

Subhotosh Khan

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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......
 

Subhotosh Khan

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Mark took you out of your misery - but please work through the intermediate steps.

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

Cubist

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

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

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

and from there it's simple!
 

MarkFL

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Mark took you out of your misery - but please work through the intermediate steps.

It will help you to build "your character" (according to my UG calculus professor).
It appeared to me that the OP was actually already at the doorstep in the first attached image, having gotten to this point:

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

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. :)
 
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