Please help
Show that lim((n+1)/(n-1))^n=e^2 when n-infinity?

Question

Please help 
Show that lim((n+1)/(n-1))^n=e^2 when n->infinity?

whpalmer4 · Answer

You need to apply a little trick.  Remember that the natural log of an exponential is the exponent times the natural log of the base:

$$\ln x^n = n\ln x$$

Use that to convert your limit to a power of $e$.  Then use a substitution to transform the fraction into something that will go to 0/0 and use L'Hopital's rule.

anonymous · Answer

it's still hard for me to understand it., please explain more

whpalmer4 · Answer

If we take the natural log of our expression inside the limit, we get $$n \ln(\frac{n+1}{n-1})$$We still can't take the limit of that directly, but we can if we make a substitution to turn it into a quotient where both numerator and denominator will go to 0 as n goes to infinity.  Let m = 1/n and our expression becomes just such a beast.  Use L’Hopital's rule to take the limit of that and the answer should be 2.  Finally, because we took the natural log of our expression, we need to take the natural anti-log of the whole thing to preserve the correct value, and that will give us the desired answer.

Give it a try, and I'll check back when I get back to my computer.

anonymous · Answer

I've been using L'Hopital's rule .... but I can not get the right answer. I do not know how to reach the answer. it will stop. I do not understand what you mean by Let m = 1 / n I feel stupid.

whpalmer4 · Answer

What I was proposing was that you rewrite the expression in terms of $m$ by substituting  $m=1/n$ or equivalently $n = 1/m$

Then you would have $$n \ln (\frac{n+1}{n-1}) = \frac{1}{m}*\ln(\frac{\frac{1}{m} + 1}{\frac{1}{m}-1}) =\frac{\ln(\frac{\frac{1}{m} + 1}{\frac{1}{m}-1})}{m}$$ and the limit changes from $$n\rightarrow\infty$$to$$m\rightarrow0$$ because of the reciprocal relationship between $n$ and $m$.  That gives us

$$\lim_{m\rightarrow0}\frac{\ln(\frac{\frac{1}{m} + 1}{\frac{1}{m}-1})}{m}$$

Can you apply L'Hopital to that?

anonymous · Answer

Thank you.
derivative of ln ((((1 / m) +1) / ((1 / m) -1))))/m is 2.
 what should I do next

whpalmer4 · Answer

Well, when we took the log, we turned our problem from 

lim (stuff)^n into e ^ lim n ln (stuff) 
If the limit of n ln (stuff) = 2, then we have shown that the limit of the original thing is e^2...

anonymous · Answer

Thanks, I appreciate your help. I have another proplem, hope you want to help me with it.

Calculate the limit of

 $${Xn}=\sin \sqrt{n+1}-\sin \sqrt{n} ,            n \rightarrow \infty$$

Thanks again

whpalmer4 · Answer

Well, I can do a hand-waving answer to that, but I can't do it with any rigor.  Might be better off posting it as a separate question in the hopes of getting someone else who knows the proper approach.

anonymous · Answer

I have done but I have not received good response. thanks