Two Solved Exercises

In my other blog I received a three challenges regarding the solution of two exercises. Since I think the comments section of m blog isn't really the place for me to post a solution I decided to post the resolutions in here.


As a matter of fact I'm kind of opening an exception that I'm quite sure I won't open many times in the future. My other blog isn't about people sending problems for me to solve (neither is this one by the way). It is about me posting the contents of my physics degree online so that other people can see what physics is about when done from the bottom up. But since I'm sure that the person that send me those exercises is perfectly able to solving them himself I'm doing this this time.


Anyway, enough with the explanations and let's get our hands dirty,because that's the only way for us to learn in Physics and Mathematics.

— 1. Exercise —

$ \displaystyle \lim \left( \dfrac{a}{1+|a|} \right)^n $

— 1.1. First Resolution —

$ {\left| \dfrac{a}{1+|a|} \right|= \dfrac{|a|}{|1+|a||}=\dfrac{|a|}{1+|a|} }$

Since $ {1+|a|>0}$ for all values of $ {a}$.


Now $ {\dfrac{|a|}{1+|a|}<1}$ since the numerator is less than the denominator. From this it follows that:

$ \displaystyle \lim \left( \dfrac{a}{1+|a|} \right)^n=0 $

— 1.2. Second Resolution —

First let us suppose that it is $ {a=0}$. In this case it is $ { \lim \left( \dfrac{a}{1+|a|} \right)^n=0 }$ trivially.


Now let us suppose that it is $ {a > 0}$

$ {\begin{aligned} \lim \left( \dfrac{a}{1+|a|}\right)^n &=\lim \left( \dfrac{1+|a|}{a}\right)^{-n}\\ &=\lim \left( \dfrac{1+a}{a}\right)^{-n} \\ &=\lim \left( \dfrac{1}{a}+ 1 \right)^{-n} \\ &= 0 \end{aligned}}$


Where the last equality follows from the fact that $ { \dfrac{1}{a} + 1 > 1 }$.


Finally let us suppose that $ {a < 0}$ (hence $ { a = - |a| }$)


$ {\begin{aligned} \lim \left( \dfrac{1-a}{a}\right)^{-n} &=\lim \left( \dfrac{1}{a}-1 \right)^{-n}\\ &=- \lim \left( \dfrac{1}{|a|} +1 \right)^{-n} \\ &= 0 \end{aligned}}$

Where the last equality follows from the previous result.

Thus it always is $ {\lim \left( \dfrac{a}{1+|a|} \right)^n = 0 }$

— 2. Exercise —

$ \displaystyle \lim \sqrt[n]{ \dfrac{(3n)!}{(n!)^3} } $

$ {\begin{aligned} \lim \exp \left[\log \left(\sqrt[n]{ \dfrac{(3n)!}{(n!)^3} }\right) \right] &=\lim \exp \left[ \dfrac{1}{n}\log \left( \dfrac{(3n)!}{(n!)^3}\right) \right]\\ &=\lim \exp \left[ \dfrac{1}{n}(\log (3n)!-3 \log n!) \right]\\ &=\exp \left[ \lim \dfrac{1}{n}(\log (3n)!-3 \log n!) \right] \end{aligned}}$


Now we'll calculate $ { \lim \dfrac{1}{n}( \log (3n)! -3 \log n! ) }$ using Stirling's approximation and neglecting the $ { O (\log n) }$ terms since $ {\lim O (\log n)/n=0}$.


$ {\begin{aligned} \lim \dfrac{1}{n} (\log (3n)! -3 \log n!) &=\lim \dfrac{1}{n} (3n \log 3n -3n -3 ( n \log n -n ) )\\ &=\lim \dfrac{1}{n} (3n \log 3 + 3n \log n -3n -3n \log n +3n)\\ &=\lim \dfrac{3n \log 3}{n}\\ &=3\log 3 \end{aligned}}$

Thus $ {\lim \sqrt[n]{ \dfrac{(3n)!}{(n!)^3} } = \exp (3 \log 3 ) = 3^3=27 }$