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 $n^n$I 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.