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

[Quinzio]

Homework Statement

\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

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 n^n, since n^n>n!>a^n>...
I should make passages, but this is quite clear to me.

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

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

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

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

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

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

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

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

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

\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

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
\lim_{x \to 0} \left(x - \sin x \right)^{\frac{1}{x}} = \left(\frac{x^3}{6}+ o(x^3) \right)^{\frac{1}{x}}

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

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

I believe this is \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) ,

making the final result

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

(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.