Limit Manipulation: Solving for Infinity in Tricky Limits | Homework Help

[Quinzio]

Homework Statement

[tex]\lim_{n \to +\infty} \left (7(6^{\frac{1}{3}}n)^{3n}-7^{3n}+(n+1)!\right) \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n [/tex]

Homework Equations

Limits manipulation

The Attempt at a Solution

Ok, in the first parenthesis, I have clear (I hope so) that the biggest term is the one containing [itex]n^n[/itex], since [itex]n^n>n!>a^n>...[/itex]
I should make passages, but this is quite clear to me.

The hard part is that
[tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n [/tex]

I am tempted to replace [itex]x = \frac{1}{n}[/itex]
so that I obtain

[tex]\lim_{x \to 0} \left(x - \sin x \right)^{\frac{1}{x}} [/tex]

Now, remembering that for [itex]x \to 0[/itex]
[tex]\sin x = x - \frac{x^3}{6}+ o(x^3)[/tex]
I'll write
[tex]\lim_{x \to 0} \left(x - \sin x \right)^{\frac{1}{x}} = \left(\frac{x^3}{6}+ o(x^3) \right)^{\frac{1}{x}} [/tex]

Let me forget the rest [itex]o(x^3)[/itex], and go back to [itex]n[/itex]
and write that

[tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n = \frac{1}{6n^{3n}}[/tex]
This should hold true, as [itex]n \to +\infty[/itex]

Now, I go back to the first part of the limit

[tex]\lim_{n \to +\infty} \left (7(6^{\frac{1}{3}}n)^{3n}-7^{3n}+(n+1)!\right) [/tex]
neglecting the parts that are lower oder of infinity wrt to [itex]n^n[/itex]I can write it as
[tex]\lim_{n \to +\infty} \left (7(6^{\frac{1}{3}}n)^{3n}\right) [/tex]
[tex]\lim_{n \to +\infty} \left (7(6^{n})n^{3n}\right) [/tex]

If I divide it by the term coming from the other expression I get

[tex]\lim_{n \to +\infty} \frac {\left (7(6^{n})(n^{3n})\right)}{6n^{3n}} = \lim_{n \to +\infty} \left (\frac{7}{6}(6^{n})\right) = +\infty [/tex]

I should have finally found that all that stuff goes to infinity.

But I'm not really sure of the passages... anyone can kindly confirm ?

 

[dynamicsolo]

I'll write
[tex]\lim_{x \to 0} \left(x - \sin x \right)^{\frac{1}{x}} = \left(\frac{x^3}{6}+ o(x^3) \right)^{\frac{1}{x}} [/tex]

Let me forget the rest [itex]o(x^3)[/itex], and go back to [itex]n[/itex]
and write that

[tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n = \frac{1}{6n^{3n}}[/tex]
This should hold true, as [itex]n \to +\infty[/itex]

I believe this is [tex]\lim_{n \to +\infty} \left(\frac{1}{n}-\sin \frac{1}{n} \right)^n = \lim_{n \to +\infty} \left(\frac{1}{6n^{3}}\right)^n = \lim_{n \to +\infty} \left(\frac{1}{6^{n}n^{3n}}\right) ,[/tex]

making the final result

[tex]\lim_{n \to +\infty} \frac {\left (7(6^{n})(n^{3n})\right)}{6^{n}n^{3n}} = 7 .[/tex]

(It took a while to spot the omission. A graph of the function confirms this limit -- although the grapher's calculator gave out before x = 60...)

 

[Quinzio]

Ah sure... poor me !

Thanks so much.