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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 ?
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